Physics Malaysian Matriculation Semester 1 Notes Complete
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Nota Padat Fizik Matrikulasi Malaysia Sesi 2013/2014TRANSCRIPT
PHYSICS CHAPTER 2
CHAPTER 2:CHAPTER 2:Kinematics of linear motionKinematics of linear motion
1
PHYSICS CHAPTER 2
Kinematics of linear motionKinematics of linear motion
2.1 Linear Motion2.2 Uniformly Accelerated Motion2.3 Free Falling Body2.4 Projectile Motion
2
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine and distinguish between and distinguish between
i.i. distance and displacement distance and displacement ii.ii. speed and velocityspeed and velocityiii.iii. instantaneous velocity, average velocity and uniform instantaneous velocity, average velocity and uniform
velocity.velocity.iv.iv. instantaneous acceleration, average acceleration and instantaneous acceleration, average acceleration and
uniform acceleration. uniform acceleration. SketchSketch graphs of displacementtime, velocitytime and graphs of displacementtime, velocitytime and
accelerationtime.accelerationtime. DetermineDetermine the distance travelled, displacement, velocity the distance travelled, displacement, velocity
and acceleration from appropriate graphs.and acceleration from appropriate graphs.
Learning Outcome:2.1 Linear Motion
PHYSICS CHAPTER 2
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2.1. Linear motion (1D)
2.1.1. Distance, d scalar quantity. is defined as the length of actual path between two pointslength of actual path between two points. For example :
The length of the path from P to Q is 25 cm.
P
Q
PHYSICS CHAPTER 2
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vector quantity is defined as the distance between initial point and final the distance between initial point and final
point in a straight linepoint in a straight line. The S.I. unit of displacement is metre (m).
Example 1:An object P moves 20 m to the east after that 10 m to the south and finally moves 30 m to west. Determine the displacement of P relative to the original position.Solution :Solution :
2.1.2 Displacement,s
N
EW
S
O
P
20 m
10 m
10 m 20 m
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The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v is defined the rate of change of distancerate of change of distance. scalar quantity. Equation:
interval timedistance of changespeed =
tdv =
PHYSICS CHAPTER 2
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is a vector quantity. The S.I. unit for velocity is m s1.
Average velocity, Average velocity, vvavav is defined as the rate of change of displacementthe rate of change of displacement. Equation:
Its direction is in the same direction of the change in same direction of the change in displacementdisplacement.
2.1.4 Velocity,v
interval timentdisplaceme of change
=avv
tsvav =
12
12av tt
ssv
=
PHYSICS CHAPTER 2
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Instantaneous velocity, Instantaneous velocity, vv is defined as the instantaneous rate of change of the instantaneous rate of change of
displacementdisplacement. Equation:
An object is moving in uniform velocitymoving in uniform velocity if
ts
0tv
=
limit
constant=dtds
dtdsv =
PHYSICS CHAPTER 2
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Therefore
Q
s
t0
s1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t1
Gradient of st graph = velocity
PHYSICS CHAPTER 2
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vector quantity The S.I. unit for acceleration is m s2.
Average acceleration, Average acceleration, aaavav is defined as the rate of change of velocitythe rate of change of velocity. Equation:
Its direction is in the same direction of motionsame direction of motion. The accelerationacceleration of an object is uniformuniform when the magnitude magnitude
of velocity changes at a constant rate and along fixed of velocity changes at a constant rate and along fixed direction.direction.
2.1.5 Acceleration, a
interval time velocityof change
=ava
12
12av tt
vva
=
tvaav =
PHYSICS CHAPTER 2
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Instantaneous acceleration, Instantaneous acceleration, aa is defined as the instantaneous rate of change of velocityinstantaneous rate of change of velocity. Equation:
An object is moving in uniform acceleration moving in uniform acceleration if
tv
0ta
=
limit
constant=dtdv
2
2
dtsd
dtdva ==
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Deceleration,Deceleration, aa is a negative accelerationnegative acceleration. The object is slowing downslowing down meaning the speed of the object speed of the object
decreases with timedecreases with time.
Therefore
v
t
Q
0
v1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t1
Gradient of vt graph = acceleration
PHYSICS CHAPTER 2
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Displacement against time graph (Displacement against time graph (stst))2.1.6 Graphical methods
s
t0
s
t0(a) Uniform velocity (b) The velocity increases with time
Gradient = constant
Gradient increases with time
(c)s
t0
Q
RP
The direction of velocity is changing.
Gradient at point R is negative.
Gradient at point Q is zero.
The velocity is zero.
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Velocity versus time graph (Velocity versus time graph (vtvt))
The gradient at point A is positive a > 0(speeding up) The gradient at point B is zero a= 0 The gradient at point C is negative a < 0(slowing down)
t1 t2
v
t0 (a) t2t1
v
t0 (b) t1 t2
v
t0 (c)
Uniform velocityUniform acceleration
Area under the vt graph = displacement
BC
A
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From the equation of instantaneous velocity,
Therefore
dtdsv =
= vdtds= 2
1
t
tvdts
graph under the area dedsha tvs =
Simulation 2.1 Simulation 2.2 Simulation 2.3
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A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.
a. Explain qualitatively the motion of the toy train.b. Sketch a velocity (cm s1) against time (s) graph.c. Determine the average velocity for the whole journey.d. Calculate the instantaneous velocity at t = 12 s.
Example 2 :
0 2 4 6 8 10 12 14 t (s)
2
4
68
10
s (cm)
Figure 2.1Figure 2.1
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Solution :Solution :0 to 6 s :
6 to 10 s : 10 to 14 s :
b.
0 2 4 6 8 10 12 14 t (s)
0.68
1.50
v (cm s1)
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Solution :Solution :c.
d.
12
12
ttssvav
=
s 14 tos 10 from velocity average=v
12
12
ttssv
=
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A velocitytime (vt) graph in figure 2.2 shows the motion of a lift.
a. Describe qualitatively the motion of the lift.b. Sketch a graph of acceleration (m s1) against time (s).c. Determine the total distance travelled by the lift and its displacement.d. Calculate the average acceleration between 20 s to 40 s.
Example 3 :
05 10 15 20 25 30 35 t (s)
42
2
4
v (m s 1)
Figure 2.2Figure 2.2
40 45 50
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Solution :Solution :a. 0 to 5 s : Lift moves upward from rest with
acceleration of 0.4 m s2. 5 to 15 s : The velocity of the lift from 2 m s1 to
4 m s1 but the acceleration to 0.2 m s2. 15 to 20 s : Lift 20 to 25 s : Lift 25 to 30 s : Lift 30 to 35 s : Lift moves
35 to 40 s : Lift moving
40 to 50 s :
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Solution :Solution :b.
t (s)5 10 15 20 25 30 35 40 45 500
0.40.2
0.2
0.6
a (m s2)
0.6
0.8
0.8
0.4
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Solution :Solution :c. i.
05 10 15 20 25 30 35 t (s)
42
2
4
v (m s 1)
40 45 50A1
A2 A3
A4 A5
vt ofgraph under the area distance Total =54321 AAAAA ++++=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145
214105
211042
2152
21distance Total +++++++=
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Solution :Solution :c. ii.
d.
vt ofgraph under the areant Displaceme =
54321 AAAAA ++++=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145
214105
211042
2152
21ntDisplaceme +++++++=
12
12
ttvvaav
=
PHYSICS CHAPTER 2
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Figure 2.3Figure 2.3
1. Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.
a. Describe the motion of the object in 10 s.b. Sketch a graph of acceleration (m s2) against time (s) for the whole journey.c. Calculate the displacement of the object in 10 s.
ANS. : 6 mANS. : 6 m
Exercise 2.1 :
PHYSICS CHAPTER 2
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2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s.a. Sketch a velocitytime graph for the journey.b. Calculate the acceleration and the distance travelled in each part of the journey.c. Calculate the average velocity for the journey.Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11
ANS. : 0.4 m sANS. : 0.4 m s22,0 m s,0 m s22,0.267 m s,0.267 m s22, 80 m, 800 m, 120 m; , 80 m, 800 m, 120 m; 6.67 m s6.67 m s11..
Exercise 2.1 :
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform
acceleration:acceleration:
Learning Outcome:2.2 Uniformly accelerated motion
atuv +=2
21 atuts +=
asuv 222 +=
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2.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constantconstant)
acceleration is given by
where v : final velocityu : initial velocitya : uniform (constant) accelerationt : time
atuv += (1)t
uva =
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From equation (1), the velocitytime graph is shown in figure 2.4:
From the graph, The displacement after time, s = shaded area under the
graph = the area of trapezium
Hence,
velocity
0
v
utimetFigure 2.4Figure 2.4
( )tvu21s += (2)
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By substituting eq. (1) into eq. (2) thus
From eq. (1),
From eq. (2),
( )[ ]tatuus ++=21
(3)2
21 atuts +=
( ) atuv =( )
tsuv 2=+
multiply
( ) ( ) ( )attsuvuv
=+2
asuv 222 += (4)
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Notes: equations (1) (4) can be used if the motion in a straight motion in a straight
line with constant acceleration.line with constant acceleration. For a body moving at constant velocity, ( ( aa = 0) = 0) the
equations (1) and (4) become
Therefore the equations (2) and (3) can be written asuv =
vts = constant velocityconstant velocity
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A plane on a runway takes 16.2 s over a distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculatea. the speed on leaving the ground,b. the acceleration during take off.Solution :Solution :
a. Use
Example 4 :
s 2.16=t
?=v
( )tvus +=21
0=u
m 1200=s
?=a
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Solution :Solution :b. By using the equation of linear motion,
asuv 222 +=
OROR2
21 atuts +=
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A bus travelling steadily at 30 m s1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s2 in the same direction as the bus. Determinea. the time taken for the car to acquire the same velocity as the bus,b. the distance travelled by the car when it is level with the bus.Solution :Solution :
a. Given Use
Example 5 :
21 ms 2 0; ;constant s m 30 ==== ccb auv
cccc tauv +=1s m 30 == bc vv
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b.
From the diagram,
c
b
1s m 30 =bv
0=cus 0=bt s 5=bt
2s m 2 =cab
bv b
c
bv
ttb =bc ss =
bc ss =
bbcccc tvtatu =+2
21
Thereforetvs bc =
;ttb = 5= ttc
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A particle moves along horizontal line according to the equation
Where s is displacement in meters and t is time in seconds. At time, t =2.00 s, determinea. the displacement of the particle,b. Its velocity, andc. Its acceleration.Solution :Solution :a. t =2.00 s ;
Example 6 :
ttts 23 243 +=
ttts 23 243 +=
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Solution :Solution :b. Instantaneous velocity at t = 2.00 s,
Use
Thus
dtdsv =
( )tttdtdv 243 23 +=
( ) ( ) 22.0082.009 2 +=v
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Solution :Solution :c. Instantaneous acceleration at t = 2.00 s,
Use
Hence
dtdva =
( ) 82.0018 =a
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1. A speedboat moving at 30.0 m s1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of 3.50 m s2 by reducing the throttle.a. How long does it take the boat to reach the buoy?b. What is the velocity of the boat when it reaches the buoy?No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.
ANS. : 4.53 s; 14.1 m sANS. : 4.53 s; 14.1 m s11
2. An unmarked police car travelling a constant 95 km h1 is passed by a speeder traveling 140 km h1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police cars acceleration is 2.00 m s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.
ANS. : 14.4 sANS. : 14.4 s
Exercise 2.2 :
PHYSICS CHAPTER 2
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3. A car traveling 90 km h1 is 100 m behind a truck traveling 75 km h1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck.No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.
ANS. : 24 sANS. : 24 s4. A car driver, travelling in his car at a constant velocity of
8 m s1, sees a dog walking across the road 30 m ahead. The drivers reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible.
ANS. : 1.73 mANS. : 1.73 m
Exercise 2.2 :
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for free falling body. equations for free falling body.
For For upward and downwardupward and downward motion, use motion, useaa = = gg = = 9.81 m s9.81 m s22
Learning Outcome:2.3 Free falling body
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2.3. Free falling body is defined as the vertical motion of a body at constant the vertical motion of a body at constant
acceleration, acceleration, gg under gravitational field under gravitational field without air without air resistanceresistance..
In the earths gravitational field, the constant acceleration known as acceleration due to gravityacceleration due to gravity or freefall freefall
accelerationacceleration or gravitational accelerationgravitational acceleration. the value is gg = 9.81 m s= 9.81 m s22 the direction is towards the centre of the earth towards the centre of the earth
(downward).(downward). Note:
In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air assumption made is ignore the air resistanceresistance.
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Sign convention:
Table 2.1 shows the equations of linear motion and freely falling bodies.
Table 2.1Table 2.1
Linear motion Freely falling bodiesatuv += gtuv =
as2uv 22 += gs2uv 22 =2at
21uts += 2gt
21uts =
+
 +

From the sign convention thus,
ga =
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An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in figure 2.5.
Assuming air resistance is negligible, the acceleration of the ball, a = g when the ball moves upward and its velocity velocity decreases to zerodecreases to zero when the ball reaches the maximum maximum height, height, HH.
H
uv
velocity = 0
Figure 2.5Figure 2.5
uv =
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The graphs in figure 2.6 show the motion of the ball moves up and down.
Derivation of equationsDerivation of equations At the maximum height or
displacement, H where t = t1, its velocity,
hence
therefore the time taken for the ball reaches H,
Figure 2.6Figure 2.6
t0
vu
u
t1 2t1
t0
a
g
t1 2t1
t
s
0
H
t1 2t1
v =0
gtuv =1gtu =0
0=v
gut1 =
Simulation 2.4
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To calculate the maximum height or displacement, H:use either
maximum height,
Another form of freely falling bodies expressions are
211 gtuts 2
1=
gsuv 22 2=Where s = H
gHu 20 2 =
OROR
guH2
2
=
gtuv =gsuv 222 =
2
21 gtuts =
gtuv yy =yyy gsuv 2
22=
2
21 gttus yy =
PHYSICS CHAPTER 2
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A ball is thrown from the top of a building is given an initial velocity of 10.0 m s1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculatea. the maximum height of the stone from point A.b. the time taken from point A to C.c. the time taken from point A to D.d. the velocity of the stone when it reaches point D.(Given g = 9.81 m s2)
Example 7 :
A
B
C
D
u =10.0 m s1
30.0 m
Figure 2.7Figure 2.7
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Solution :Solution :a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus
b. From point A to C, the vertical displacement, sy= 0 m thus
y2y
2y gsuv 2=
2yy gttus 2
1=
A
B
C
D
u
30.0 m
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Solution :Solution :c. From point A to D, the vertical displacement, sy= 30.0 m thus
By using
2yy gttus 2
1=
A
B
C
D
u
30.0 m2a
4acbb 2 =t
ORTime dont Time dont have have negative negative value.value.
a b c
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Solution :Solution :d. Time taken from A to D is t = 3.69 s thus
From A to D, sy = 30.0 m
Therefore the balls velocity at D is
A
B
C
D
u
30.0 m
gtuv yy =( ) ( ) ( )3.699.8110.0 =yvOR
y2
y2
y gsuv 2=( ) ( )( )30.09.81210.0 = 22yv
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A book is dropped 150 m from the ground. Determinea. the time taken for the book reaches the ground.b. the velocity of the book when it reaches the ground.(given g = 9.81 m s2)Solution :Solution :
a. The vertical displacement issy = 150 m
Hence
Example 8 :
uy = 0 m s1
150 mm 150=ys
2yy gttus 2
1=
PHYSICS CHAPTER 2
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Solution :Solution :b. The books velocity is given by
Therefore the books velocity is
gtuv yy =
OR
y2
y2
y gsuv 2=m 150=ys
0=yu
?=yv
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1. A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground,b. the balls speed when it reaches the ground.
ANS. : 1.79 s; 25.6 m sANS. : 1.79 s; 25.6 m s11
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.
From what height above the top of the windows did the stone fall?
ANS. : 1.75 mANS. : 1.75 m
Exercise 2.3 :
m 2.2
Figure 2.8Figure 2.8
to travel this distance took 0.30 s
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for projectile, equations for projectile,
CalculateCalculate time of flight, maximum height, range, time of flight, maximum height, range, maximum range, instantaneous position and velocity.maximum range, instantaneous position and velocity.
Learning Outcome:2.4 Projectile motion
uux cos=uu y sin=
0=xagay =
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2.4. Projectile motion A projectile motion consists of two components:
vertical component (ycomp.) motion under constant acceleration, ay= g
horizontal component (xcomp.) motion with constant velocity thus ax= 0
The path followed by a projectile is called trajectory is shown in figure 2.9.
v
u
sx= R
sy=H
ux
v2yuy
v1x
v1y
v2x
v11
v22
t1 t2
B
A
P Q
C
y
xFigure 2.9Figure 2.9
Simulation 2.5
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From figure 2.9, The xcomponent of velocityxcomponent of velocity along AC (horizontal) at any
point is constant,constant,
The ycomponent (vertical) of velocity variesycomponent (vertical) of velocity varies from one point to another point along AC.but the ycomponent of the initial velocity is given by
uux cos=
uu y sin=
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Table 2.2 shows the x and ycomponents, magnitude and direction of velocities at points P and Q.
Velocity Point P Point Q
xcomp.
ycomp.
magnitude
direction
11 gtuv yy =uuv xx1 cos==
22 gtuv yy =uuv xx2 cos==
( ) ( )2y12x11 vvv +=
=
x1
y111 v
v tan
( ) ( )2y22x22 vvv +=
=
x2
y212 v
v tan
Table 2.2Table 2.2
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The ball reaches the highest point at point B at velocity, v where xcomponent of the velocity, ycomponent of the velocity, ycomponent of the displacement,
Use
2.4.1 Maximum height, H
uuvv xx cos===0=yv
yyy gsuv 222
=
( ) gHu 2sin0 2 =
guH
2sin 22
=
Hsy =
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At maximum height, H Time, t = t and vy= 0
Use
2.4.2 Time taken to reach maximum height, t
gtuv yy =( ) 'sin0 tgu =
gut sin' =
2.4.3 Flight time, t (from point A to point C)'2 tt =
gut sin2=
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Since the xcomponent for velocity along AC is constant hence
From the displacement formula with uniform velocity, thus the xcomponent of displacement along AC is
2.4.4 Horizontal range, R and value of R maximum
tus xx =
cosuvu xx ==
( ) ( )tuR = cos( )
=
guuR sin2cos
( ) cossin22guR =
and Rsx =
PHYSICS CHAPTER 2
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From the trigonometry identity,
thus
The value of R maximum when = = 4545 and sin 2sin 2 = = 11 therefore
cossin22sin =
2sin2
guR =
guR
2
max =
Simulation 2.6
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Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.
Horizontal component along path AB.
Vertical component along path AB.
2.4.5 Horizontal projectile
h
xA B
u u
vxv
yv
Figure 2.10Figure 2.10
constant velocity, === xx vuuxsx = nt,displaceme
0u y = velocity,initialhsy = nt,displaceme
Simulation 2.7
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Time taken for the ball to reach the floor (point B), Time taken for the ball to reach the floor (point B), tt By using the equation of freely falling bodies,
Horizontal displacement, Horizontal displacement, xx Use condition below :
2yy gttus 2
1=
2gt0h21
=
ght 2=
The time taken for the ball free fall to point A
The time taken for the ball to reach point B=
(Refer to figure 2.11)
Figure 2.11Figure 2.11
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Since the xcomponent of velocity along AB is constant, thus the horizontal displacement, x
Note : In solving any calculation problem about projectile motion,
the air resistance is negligibleair resistance is negligible.
tus xx =
=
ghux 2
and xsx =
PHYSICS CHAPTER 2
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Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s1 and makes an angle, = 60.0 to the horizontal. Determinea. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.
Example 9 :
Figure 2.12Figure 2.12 xO
u
= 60.0
y
R
H
v2y
v1x
v1y v2xQv1
P
v2
PHYSICS CHAPTER 2
65
b. the time taken for the ball reaches the maximum height, H and calculate the value of H.
c. the horizontal range, Rd. the magnitude and direction of its velocity when the ball
reaches the ground (point P).e. the position of the ball, and the magnitude and direction of its
velocity at point Q if the ball was hit from a flattopped hill with the time at point Q is 45.0 s.
(given g = 9.81 m s2)Solution :Solution :The component of Initial velocity :
PHYSICS CHAPTER 2
66
Solution :Solution :a. i. position of the ball when t = 2.0 s ,
Horizontal component :
Vertical component :
therefore the position of the ball is
2yy gttus 2
1=
tus xx =
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Solution :Solution :a. ii. magnitude and direction of balls velocity at t = 2.0 s ,
Horizontal component :
Vertical component :
Magnitude,
Direction,
gtuv yy =
1xx uv
== s m 100
( ) ( ) 2222 153100 +=+= yx vvv
=
=
100153tantan 11
x
y
vv
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Solution :Solution :b. i. At the maximum height, H :
Thus the time taken to reach maximum height is given by
ii. Apply
gtuv yy =
0=yv
gttus yy 21
=
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Solution :Solution :c. Flight time = 2(the time taken to reach the maximum height)
Hence the horizontal range, R is
d. When the ball reaches point P thusThe velocity of the ball at point P,Horizontal component:Vertical component:
( )17.62=t
tus xx =
11 s m 100
== xx uv
0=ys
gtuv yy =1
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Solution :Solution :Magnitude,
Direction,
therefore the direction of balls velocity is
e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s,
Horizontal component :
( ) ( ) 2221211 172100 +=+= yx vvv
=
=
100172tantan 1
1
11
x
y
vv
300= from positive xaxis anticlockwisefrom positive xaxis anticlockwise
tus xx =
PHYSICS CHAPTER 2
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Solution :Solution :Vertical component :
therefore the position of the ball is (4500 m, (4500 m, 2148 m)2148 m)e. ii. magnitude and direction of balls velocity at t = 45.0 s ,
Horizontal component :
Vertical component :
2yy gttus 2
1=
gtuv yy =2
12 s m 100
== xx uv
PHYSICS CHAPTER 2
72
Solution :Solution :Magnitude,
Direction,
therefore the direction of balls velocity is
( ) ( ) 222 269100 +=v22
222 yx vvv +=
=
x
y
vv
2
21tan
PHYSICS CHAPTER 2
73
A transport plane travelling at a constant velocity of 50 m s1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculatea. the flight time of the parcel,b. the velocity of impact of the parcel,c. the distance from X to the point of impact.(given g = 9.81 m s2)Solution :Solution :
Example 10 :
300 m
d
1s m 50 =u
X
PHYSICS CHAPTER 2
74
Solution :Solution :The parcels velocity = planes velocity
thus
a. The vertical displacement is given by
Thus the flight time of the parcel is
1s m 50 == uux
1s m 50 =uand 1s m 0 =yu
2
21 gttus yy =
PHYSICS CHAPTER 2
75
Solution :Solution :b. The components of velocity of impact of the parcel:
Horizontal component:Vertical component:
Magnitude,
Direction,
therefore the direction of parcels velocity is
1s m 50 == xx uv
( ) ( )7.829.810 =yvgtuv yy =
=
=
506.77tantan 11
x
y
vv
( ) ( ) 2222 6.7750 +=+= yx vvv
PHYSICS CHAPTER 2
76
Solution :Solution :c. Let the distance from X to the point of impact is d.
Thus the distance, d is given by
tus xx =
PHYSICS CHAPTER 2
77
Figure 2.13Figure 2.13
Use gravitational acceleration, g = 9.81 m s21. A basketball player who is 2.00 m tall is standing on the floor
10.0 m from the basket, as in figure 2.13. If he shoots the ball at a 40.0 angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.
ANS. : 10.7 m sANS. : 10.7 m s11
Exercise 2.4 :
PHYSICS CHAPTER 2
78
2. An apple is thrown at an angle of 30 above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s1. Calculatea. the time taken for the apple to strikes the ground,b. the distance from the foot of the building will it strikes
the ground,c. the maximum height reached by the apple from the
ground.ANS. : 4.90 s; 170 m; 40.4 mANS. : 4.90 s; 170 m; 40.4 m
3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s1 at 40 above the horizontal. How far above or below its original level will the stone strike the opposite wall?
ANS. : 10.3 m below the original level.ANS. : 10.3 m below the original level.
Exercise 2.4 :
PHYSICS CHAPTER 2
79
THE ENDNext Chapter
CHAPTER 3 :Force, Momentum and Impulse
PHYSICS CHAPTER 3
1
MOMENTUM AND IMPULSE
CHAPTER 3
PHYSICS CHAPTER 3
3.0 MOMENTUM AND IMPULSE
3.1 Momentum and impulse
3.2 Conservation of linear momentum
2
PHYSICS CHAPTER 3
3
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
DefineDefine momentum.
DefineDefine impulse and use Ft graph to determine impulse
Use Use
Learning Outcome:3.1 Momentum and impulse
PHYSICS CHAPTER 3
4
3.1.1 Linear momentum,
is defined as the product between mass and velocitythe product between mass and velocity. is a vector quantity. Equation :
The S.I. unit of linear momentum is kg m skg m s11. The direction of the momentumdirection of the momentum is the samesame as the direction direction
of the velocityof the velocity. It can be resolve into vertical (y) component and horizontal (x)
component.
p
vmp =
xp
pyp
mvppx coscos ==mvppy sinsin ==
PHYSICS CHAPTER 3
5
3.1.2 Impulse, Let a single constant force, constant force, FF acts on an object in a short time
interval (collision), thus the Newtons 2nd law can be written as
is defined as the product of a force, the product of a force, FF and the time, and the time, tt OR the change of momentumthe change of momentum.
is a vector quantityvector quantity whose directiondirection is the samesame as the constant forceconstant force on the object.
J
constant === dtpdFF
12 pppddtFJ
===
momentum final: 2p
wheremomentum initial: 1p
force impulsive :F
PHYSICS CHAPTER 3
6
The S.I. unit of impulse is N sN s or kg m skg m s11. If the forceforce acts on the object is not constantnot constant then
Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :
dtFdtFJ avt
t
== 2
1
where force impulsive average :avF
( ) ( )xxx1x2xavx uvmppdtFJ ===( ) ( )yyy1y2yavy uvmppdtFJ ===( ) ( )zzz1z2zavz uvmppdtFJ ===
consider 2D consider 2D collision onlycollision only
PHYSICS CHAPTER 3
7
When two objects in collision, the impulsive force, F against time, t graph is given by the Figure 3.20.
1t 2tFigure 3.20Figure 3.20 t0
F
Shaded area under the Ft graph = impulse
Picture 3.1
Picture 3.2
Picture 3.3
PHYSICS CHAPTER 3
8
A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s1 and it bounces off with a speed of 70 m s1 in the opposite direction.a. Calculate the magnitude of impulse delivered to the ball by the wall,b. If the ball is in contact with the wall for 10 ms, determine the magnitude of average force exerted by the wall on the ball.Solution :Solution :
Example 3.1 :
Wall (2)11
1s m 100 =1u
111s m 70 =1v
022 == uv
kg 0.201 =m
PHYSICS CHAPTER 3
9
Solution :Solution :a. From the equation of impulse that the force is constant,
Therefore the magnitude of the impulse is 34 N s34 N s.
b. Given the contact time,
12 ppdpJ ==( )111 uvmJ =
dtFJ av=
PHYSICS CHAPTER 3
10
An estimated forcetime curve for a tennis ball of mass 60.0 g struck by a racket is shown in Figure 3.21. Determinea. the impulse delivered to the ball,b. the speed of the ball after being struck, assuming the ball is being served so it is nearly at rest initially.
Example 3.2 :
0.2 1.8 ( )mst0
( )kNF
1.0
18
Figure 3.21Figure 3.21
PHYSICS CHAPTER 3
11
Solution :Solution :a. From the forcetime graph,
b. Given the balls initial speed,
graph under the area tFJ =
0=u( )uvmdpJ ==
kg 1060.0 3=m
PHYSICS CHAPTER 3
12
1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m.a. Calculate the impulse delivered to the ball during impact.b. If the ball is in contact with the slab for 2.00 ms, determine the average force on the ball during impact.
ANS. : 0.47 N s; 237. 1 NANS. : 0.47 N s; 237. 1 N2. A golf ball (m = 46.0 g) is struck with a force that makes an
angle of 45 with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.)
ANS. : 293 NANS. : 293 N
Exercise 3.1 :
PHYSICS CHAPTER 3
13
Figure 3.22Figure 3.22
3.
A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s1 strikes a wall at a 45 angle and rebounds with the same speed at 45 as shown in Figure 3.22. Calculate the impulse given by the wall.
ANS. : 2.4 N s to the left or ANS. : 2.4 N s to the left or 2.4 N s2.4 N s
Exercise 3.1 :
PHYSICS CHAPTER 3
14
3.2 Conservation of linear momentum
14
3.2.1 Principle of conservation of linear momentum states In an isolated (closed) system, the total momentum In an isolated (closed) system, the total momentum
of that system is constantof that system is constant. ORWhen the net external force on a system is zero, the total When the net external force on a system is zero, the total momentum of that system is constantmomentum of that system is constant.
In a Closed system,
From the Newtons second law, thus
0 == dtpdF
0= F
0=pd
PHYSICS CHAPTER 3
15
According to the principle of conservation of linear momentum, we obtain
OR
The total of initial momentum = the total of final momentumThe total of initial momentum = the total of final momentum
= fi pp
constant=p
constant= xpconstant= yp
Therefore then
PHYSICS CHAPTER 3
16
Linear momentum in one dimension collisionLinear momentum in one dimension collisionExample 3.3 :
Figure 3.14 shows an object A of mass 200 g collides headon with object B of mass 100 g. After the collision, B moves at a speed of 2 m s1 to the left. Determine the velocity of A after Collision.SolutionSolution : :
1s m 6 =Au
AB
1s m 3 =Bu
Figure 3.14Figure 3.14
= fi pp 1s m 6 ;kg 0.100 ;kg 0.200 === ABA umm
11 s m 2 ;s m 3 == BB vu
PHYSICS CHAPTER 3
17
Linear momentum in two dimension collisionLinear momentum in two dimension collisionExample 3.4 :
A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the tennis ball is deflected 50 from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s1 and v1 = 4 m s1. Calculate the magnitude and direction of soccer ball after the collision.
Figure 3.15Figure 3.15
1u
Before collision After collision
m1 m2
m1 1v
50
Simulation 3.3
PHYSICS CHAPTER 3
18
Solution :Solution :
From the principle of conservation of linear momentum,
The xcomponent of linear momentum,
= fi pp
x22x11x22x11 vmvmumum +=+
;s m 20 ;kg 0.900 ;kg 0.250 1=== 121 umm0 ;s m 4 ;0 1 5vu 112 ===
= fxix pp
PHYSICS CHAPTER 3
19
Solution :Solution :The ycomponent of linear momentum,
Magnitude of the soccer ball,
Direction of the soccer ball,
yy vmvm 22110 += = fyiy pp
( ) ( )2y22x22 vvv +=
=
=
4.840.851tantan 1
x2
y212 v
v
PHYSICS CHAPTER 3
20
1. An object P of mass 4 kg moving with a velocity 4 m s1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s1 towards it.a. Determine the total momentum before collision.b. If P immediately stop after the collision, calculate the final velocity of Q.c. If the two objects stick together after the collision, calculate
the final velocity of both objects.ANS. : 10 kg m sANS. : 10 kg m s11; 5 m s; 5 m s11 to the right; 1.7 m s to the right; 1.7 m s11 to the right to the right2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his
hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s1. Determinea. the recoil velocity of the rifle,b. the final momentum of the system.
ANS. : ANS. : 0.5 m s0.5 m s11; U think.; U think.
Exercise 3.2 :
PHYSICS CHAPTER 3
21
3.
In Figure 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s1 and 1.40 m s1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determinea. the speed of the bullet immediately after it emerges from the first block andb. the initial speed of the bullet.ANS. : 721 m sANS. : 721 m s11; 937.4 m s; 937.4 m s11
Figure 3.16Figure 3.16
1.20 kg 1.80 kg
0.630 m s1 1.40 m s1Before
After
PHYSICS CHAPTER 3
22
Figure 3.17Figure 3.17
4. A ball moving with a speed of 17 m s1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45 from its original direction, and the struck ball moves off at 30 from the original direction as shown in Figure 3.17. Calculate the speed of each ball after the collision.
ANS. : 8.80 m sANS. : 8.80 m s 1 1; 12.4 m s; 12.4 m s11
Exercise 3.2 :
PHYSICS CHAPTER 3
23
3.2.2 Collision
is defined as an isolated event in which two or more bodies an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each (the colliding bodies) exert relatively strong forces on each other for a relatively short timeother for a relatively short time.
Two types of collisions :
Elastic collisionElastic collision
Inelastic (nonelastic) collisionInelastic (nonelastic) collision
PHYSICS CHAPTER 3
24
Elastic collisionElastic collision is defined as one in which the total kinetic energy (as well as one in which the total kinetic energy (as well as
total momentum) of the system is the same before and after total momentum) of the system is the same before and after the collisionthe collision.
Figure 3.18 shows the headon collision of two billiard balls.
11 22
Before collision
At collision
After collision
11 2222um11um
11 2222vm11vm
Figure 3.18Figure 3.18
Simulation 3.4
PHYSICS CHAPTER 3
25
The properties of elastic collisionproperties of elastic collision are
a. The total momentum is conservedtotal momentum is conserved.
b. The total kinetic energy is conservedtotal kinetic energy is conserved.
OR
= fi pp
= fi KK
222
211
222
211 vmvmumum 2
121
21
21
+=+
PHYSICS CHAPTER 3
26
Inelastic (nonelastic) collisionInelastic (nonelastic) collision is defined as one in which the total kinetic energy of the one in which the total kinetic energy of the
system is not the same before and after the collision (even system is not the same before and after the collision (even though the total momentum of the system is conserved)though the total momentum of the system is conserved).
Figure 3.19 shows the model of a completely inelastic completely inelastic collisioncollision of two billiard balls.
11 22At collision
After collision (stick together)
11 22v
Figure 3.19Figure 3.19
Before collision 11 2211um
0=2u
2m
Simulation 3.5
PHYSICS CHAPTER 3
27
Caution: Not allNot all the inelastic collision is stick togetherstick together. In fact, inelastic collisions include many situationsmany situations in which
the bodies do not stickbodies do not stick
The properties of inelastic collisionproperties of inelastic collision area. The total momentum is conservedtotal momentum is conserved.
b. The total kinetic energy is not conservedtotal kinetic energy is not conserved because some of the energy is converted to internal energyinternal energy and some of it is transferred away by means of sound or heatsound or heat. But the total total energy is conservedenergy is conserved.
OR
= fi pp
= fi EE energy losses += fi KK
PHYSICS CHAPTER 3
28
Ball A of mass 400 g and velocity 4 m s1 collides with ball B of mass 600 g and velocity 10 m s1. After collision, A and B will move together. Determine the final velocity of both balls if A and B moves in the opposite direction initiallySolution : Solution : mmAA = 0.4 kg, u = 0.4 kg, uAA = 4 m s = 4 m s1 1 , m, mB B = 0.6 kg, u= 0.6 kg, uB B = 10 m s= 10 m s11, , inelastic collisioninelastic collision
By using the principle of conservation of linear momentum, thus
Example 3.5 :
Before collision AA BBBuAu
After collision AA BB?=v
= fi pp vmmumum BABBAA )( +=+
PHYSICS CHAPTER 3
29
Solution :Solution :
Final velocity of both balls is  4.4 m s 1
BA
BBAA
mmumumv
+
+=
PHYSICS CHAPTER 3
30
A ball A of mass 1 kg moving at a velocity of 4 m s1 collides with ball B of mass 2 kg which at rest. Calculate the velocity of both balls after collision if the collision is an elastic collision. Solution :Solution :
Given mA = 1 kg, uA = 4 m s1, mB = 2 kg, uB = 0 m s1, elastic collision
Example 3.6 :
Before collision AA BB
10 = msuBAu
After collision AA BB?=Bv?=Av
PHYSICS CHAPTER 3
31
Solution :Solution :Apply principle of conservation of momentum,
Apply principle of conservation of kinetic energy,
BBAABBAA vmvmumum +=+)(2)(1)0(2)4(1 BA vv +=+
222 )(21)(
21)(
21
BBAAAA vmvmum +=
222 )()()( BBAAAA vmvmum +=
1B s m v2 4
=Av ..(1)
PHYSICS CHAPTER 3
32
222 )(2)(1)4(1 BA vv +=Solution :Solution :
2216 BA vv += ..(2)Substitute equation (1) into equation (2)
22 2)24(16 BB vv +=22 2)2(416 BB vv +=
22)2(28 BB vv +=
PHYSICS CHAPTER 3
33
Substitute into equation (1),
238 BB vv =
167.2 = msvB
(2.67) 2 4=Av
Solution :Solution :
PHYSICS CHAPTER 3
34
THE ENDNext Chapter
CHAPTER 4 :Forces
PHYSICS CHAPTER 4
1
CHAPTER 4:
FORCES
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PHYSICS CHAPTER 4
4. FORCES
4.1 Basic of Forces and Free Body Diagram
4.2 Newtons Laws of Motion
2
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PHYSICS CHAPTER 4
3
At the end of this chapter, students should be able to:
Identify the forces acting on a body in different situations.
Weight
Tension
Normal force
Friction
Determine weight, static friction and kinetic friction
Draw free body diagram
Determine the resultant force
Learning Outcome:
4.1 Basic of Forces and Free Body Diagram
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PHYSICS CHAPTER 4
4.1 Basic of Forces and Free Body DiagramWeight,
is defined as the force exerted on a body under gravitational
field.
It is a vector quantity.
It is dependant on where it is measured, because the value of gvaries at different localities on the earths surface.
It always directed toward the centre of the earth or in the same
direction of acceleration due to gravity, g.
The S.I. unit is kg m s2 or Newton (N).
Equation:
gmW
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PHYSICS CHAPTER 4
5
Tension, T
The tension force is the force that is transmitted through a
string, rope, cable or wire when it is pulled tight by forces
acting from opposite ends. The tension force is directed
along the length of the wire and pulls equally on the objects
on the opposite ends of the wire.
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Figure 4.1
PHYSICS CHAPTER 4
66
Normal (reaction) force,
is defined as a reaction force that exerted by the surface to
an object interact with it and the direction always
perpendicular to the surface.
An object lies at rest on a flat horizontal surface as shown in
Figure 4.2.
Ror N
N
gmW
0mgNFy
mgN
ThereforeFigure 4.2
Action: weight of an object is exerted on the
horizontal surface
Reaction: surface is exerted a force, N on the object
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A free body diagram is defined as a diagram showing the chosen body by itself, with vectors
drawn to show the magnitude and directions of all the forces applied to the body by the other bodies
that interact with it.
PHYSICS CHAPTER 4
7
Friction
is defined as a force that resists the motion of one surface relative to another with which it is in contact.
is independent of the area of contact between the two surfaces..
is directly proportional to the reaction force.
OR
Coefficient of friction, is defined as the ratio between frictional force to reaction
force.
OR
is dimensionless and depends on the nature of the surfaces.
Nf
Nf force frictional:f
friction oft coefficien :
forcereaction : N
where
N
f
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PHYSICS CHAPTER 4
8
There are three types of frictional force :
Static, fs (frictional force act on the object before its move)
Kinetic, fk (frictional force act on the object when its move)
Rolling, fr (frictional force act on the object when its rolling)
Caution:
The direction of the frictional force exerted by a surface
on an object is always in the opposite direction of the
motion.
The frictional and the reaction forces are always
perpendicular.
Nf kk
Nf ss
Nf rr
skr fff where
thus skr
Can be ignored
Simulation 4.1
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PHYSICS CHAPTER 4
Example 4.1:
A mass is resting on a flat surface which has a normal force of
98N, with a coefficient of static friction of 0.35. What force
would it take to move the object?
9
Solution: N = 98N, s = 0.35
Nf ss
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PHYSICS CHAPTER 4
10
Example 4.2:
A 15 kg piece of wood is placed on top of another piece of
wood. There is 35N of static friction measured between them.
Determine the coefficient of static friction between the two
pieces of wood.
Solution: N = mg = 15(9.81) = 147.15 N, Fs = 35 N
N
fss
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PHYSICS CHAPTER 4
Example 4.3
A dock worker loading crates on a ship finds that a 15 kg crate,
initially at rest on a horizontal surface, requires a 50 N
horizontal force to set it in motion. However, after the crate is in
motion, a horizontal force of 30 N is required to keep it moving
with a constant speed. The acceleration of gravity is 9.8 ms2.
Find the coefficient of kinetic friction.
11
Solution:
Mass of crate = m = 15 kg
Force required to set the crate in motion = F1 = 50 N
Force required to keep the crate in moving at constant speed =
fk = 30 N
Acceleration of gravity = g = 9.81 ms2
Normal force, N = mg = =
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PHYSICS CHAPTER 4
12
Solution:
From
N
fkk
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PHYSICS CHAPTER 4
Resultant force Is defined as a single force that represents the combined
effect of two or more forces
1313
The figure above shows three forces F1, F2 and F3 acted on a
particle O. Calculate the magnitude and direction of the
resultant force on particle O.
Example 4.4:y
30o
O
)N30(2F
)N10(1F
30ox
)N40(3F
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PHYSICS CHAPTER 4
14
30o
Solution :
O
y
x
3F
30o
y3F
321 FFFFFr
yxr FFF
xxxx FFFF 321
yyyy FFFF 321
xF2
1F
2F
60o
yF2
x3F
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PHYSICS CHAPTER 4
15
Vector xcomponent ycomponent
1F
3F
2F
N 01 xF 11FF y
N 011 yF
60cos302 xFN 152 xF
60sin302 yF
N 622 yF
30cos403 xFN 34.63 xF
30sin403 yFN 203 yF
Vector
sum
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PHYSICS CHAPTER 4
16
y
xO
Solution :
The magnitude of the resultant force is
and
Its direction is 162 from positive xaxis OR 18 above negative x
axis.
22 yxr FFF
x
y
F
F 1tan
yF
xF
162
rF
18
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PHYSICS CHAPTER 4
17
1. Given three vectors P, Q and R as shown in Figure 4.3.
Calculate the resultant vector of P, Q and R.
ANS. : 49.4 m s2; 70.1 above + xaxis
Exercise 4.1:
Figure 4.3
y
x0
50 2s m 10 R
2s m 35 P
2s m 24 Q
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PHYSICS CHAPTER 4
18
At the end of this chapter, students should be able to:
State Newtons First Law
Define mass as a measure of inertia.
Define the equilibrium of a particle.
Apply Newtons First Law in equilibrium of forces
State and apply Newtons Second Law
State and apply Newtons Third Law.
Learning Outcome:
4.2 Newtons laws of motion
td
vdm
td
mdvmv
td
d
td
dpF
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PHYSICS CHAPTER 4
19
4.2 Newtons laws of motion
4.2.1 Newtons first law of motion states an object at rest will remain at rest, or continues to
move with uniform velocity in a straight line unless it is acted upon by a external forces
OR
The first law gives the idea of inertia.
0FFnett
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PHYSICS CHAPTER 4
20
4.2.2 InertiaInertia
is defined as the tendency of an object to resist any change in its state of rest or motion.
is a scalar quantity.
Mass, m
is defined as a measure of a bodys inertia.
is a scalar quantity.
The S.I. unit of mass is kilogram (kg).
The value of mass is independent of location.
If the mass of a body increases then its inertia will increase.
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PHYSICS CHAPTER 4
21
Figures 4.4a and 4.4b show the examples of real experience of inertia.
Figure 4.4
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PHYSICS CHAPTER 4
22
4.2.3 Equilibrium of a particle
is defined as the vector sum of all forces acting on a particle (point) must be zero.
The equilibrium of a particle ensures the body in translational equilibrium and its condition is given by
This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
There are two types of equilibrium of a particle. It is
Static equilibrium (v=0) body remains at rest (stationary).
Dynamic equilibrium (a=0) body moving at a uniform (constant) velocity.
0nettFF
Newtons first law of motion
0 , 0 , 0 zyx FFF
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PHYSICS CHAPTER 4
23
Problem solving strategies for equilibrium of a
particle
The following procedure is recommended when dealing with
problems involving the equilibrium of a particle:
Sketch a simple diagram of the system to help
conceptualize the problem.
Sketch a separate free body diagram for each body.
Choose a convenient coordinate axes for each body and
construct a table to resolve the forces into their
components.
Apply the condition for equilibrium of a particle in
component form :
Solve the component equations for the unknowns.
0xF 0yFand
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PHYSICS CHAPTER 4
2424
A load of 250 kg is hung by a cranes cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to the
vertical plane. If the load is in the equilibrium, calculate
a. the magnitude of the tension in the cable,
b. the magnitude of the horizontal force. (Given g =9.81 m s2)
Solution :
Example 4.5:
30
F F
Free body diagram of the load :
gm
T
yT3060
xT
kg 250m
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PHYSICS CHAPTER 4
25
Solution :
1st method :
a.
Since the load is in the equilibrium, then
Thus
b. By substituting eq. (2) into eq. (1), therefore
0xF 060cos TF
kg 250m
Force xcomponent (N) ycomponent (N)
gm
0 9.81250mg
F
F 0
T 60cosT 60sinT
2453
0F
(1)
(2) 0yF 0245360sin T
060cos2833 F
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PHYSICS CHAPTER 4
26
30
Solution :
2nd method :
a. Since the load is in the equilibrium, then a closed triangle of
forces can be sketched as shown below.
b. 30sinT
F
kg 250m
30cosT
mg
30sin2833
F
F
gm T
From the closed triangle of forces, hence
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PHYSICS CHAPTER 4
27
Calculate the magnitude and direction of a force that balance the
three forces acted at point A as shown in Figure 4.5.
Example 4.6:
N 121FN 202F
N 303F
30.055.0
45.0A
Figure 4.5
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PHYSICS CHAPTER 4
28
Solution :
To find a force to balance the three forces means the system must
be in equilibrium hence
N 30 N; 20 N; 12 321 FFF
Force xcomponent (N) ycomponent (N)
1F 55.0cos12
F
xF yF
6.88
55.0sin129.83
2F 30.0cos20
17.3
30.0sin2010.0
3F 45.0cos30
21.2
45.0sin3021.2
0 xF
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PHYSICS CHAPTER 4
29
Solution :
The magnitude of the force,
and its direction,
0 yF021.210.09.83 yF
222y2x FFF 1.3731.6
x
y1
F
F tan
31.6
1.37tan 1
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PHYSICS CHAPTER 4
30
30
A window washer pushes his scrub brush up a vertical window at
constant speed by applying a force F as shown in Figure 4.6. The brush weighs 10.0 N and the coefficient of kinetic friction is
k= 0.125. Calculate
a. the magnitude of the force F ,
b. the normal force exerted by the window on the brush.
Example 4.7:
F
50.0
Figure 4.6
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PHYSICS CHAPTER 4
31
Solution :
a. The free body diagram of the brush :
The brush moves up at constant speed (a=0) so that
Thus
0.125 ;N 10.0 kW
W
F
N
kf
constant
speed
Force xcomponent (N) ycomponent (N)
F
50.0cosF
kf
0Nk
50.0sinF
W
0 10.0
N
N 0
N0.125
0amF
50.0cosFN 0 xF (1)(2)10.00.12550.0sin NF 0 yF
50.0
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PHYSICS CHAPTER 4
32
Solution :
a. By substituting eq. (1) into eq. (2), thus
b. Therefore the normal force exerted by the window on the brush
is given by
10.050.0cos0.12550.0sin FF
50.0cosFN
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PHYSICS CHAPTER 4
33
Exercise 4.2:
Use gravitational acceleration, g = 9.81 m s2
1.
The system in Figure 5.8 is in equilibrium, with the string at the
centre exactly horizontal. Calculate
a. the tensions T1, T2 and T3.
b. the angle .ANS. : 49 N, 28 N, 57 N; 29
Figure 4.7
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PHYSICS CHAPTER 4
34
Exercise 4.2:
2.
A 20 kg ball is supported from the ceiling by a rope A. Rope B
pulls downward and to the side on the ball. If the angle of A to
the vertical is 20 and if B makes an angle of 50 to the vertical
as shown in Figure 4.8, Determine the tension in ropes A and B.
ANS. : 134 N; 300 N
Figure 4.8
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PHYSICS CHAPTER 4
35
Exercise 4.2:
3.
A block of mass 3.00 kg is pushed up against a wall by a force
P that makes a 50.0 angle with the horizontal as show in Figure 4.9. The coefficient of static friction between the block
and the wall is 0.250. Determine the possible values for the
magnitude of P that allow the block to remain stationary.
ANS. : 31.8 N; 48.6 N
Figure 4.9
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PHYSICS CHAPTER 4
36
Newtons second law of motion
states the rate of change of linear momentum of a moving body is proportional to the resultant force and is in the
same direction as the force acting on it
OR
its can be represented by
dt
pdF
where
momentumlinear in change : pd
interval time:dt
forceresultant : F
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PHYSICS CHAPTER 4
37
From the Newtons 2nd law of motion, it also can be written as
Case 1:
Object at rest or in motion with constant velocity but with
changing mass. For example : Rocket
dt
vdm
dt
dmvF
dt
pdF
dt
vmdF
dt
vdm
dt
dmvF
mvp and
0dt
vd
dt
dmvF
and
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PHYSICS CHAPTER 4
38
Case 2:
Object at rest or in motion with constant velocity and constant
mass.
Thus
dt
vdm
dt
dmvF
Newtons 1st law of motion
0 dt
pdF
constantp
0dt
dm 0
dt
vd
0F
where and
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PHYSICS CHAPTER 4
39
Case 3:
Object with constant mass but changing velocity.
The direction of the resultant force always in the same
direction of the motion or acceleration.
dt
vdm
dt
dmvF
0dt
dman
d
amF
dt
vdmF
dt
vda
and
where
objectan of mass : m
onaccelerati :a
forceresultant : F
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PHYSICS CHAPTER 4
40
Newtons 2nd law of motion restates that The acceleration of an object is directly proportional to the nett force acting on
it and inversely proportional to its mass.
OR
One newton(1 N) is defined as the amount of nett force that
gives an acceleration of one metre per second squared to a
body with a mass of one kilogramme.
OR 1 N = 1 kg m s2
Notes:
is a nett force or effective force or resultant force.
The force which causes the motion of an object.
If the forces act on an object and the object moving at
uniform acceleration (not at rest or not in the
equilibrium) hence
amFFnett
m
Fa
F
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PHYSICS CHAPTER 4
41
Newtons third law of motion states every action force has a reaction force that is equal
in magnitude but opposite in direction. For example :
When the student push on the wall it will push back with the same force. (refer to Figure 4.10)
BAAB FF
A (hand)
B (wall)
BAF
ABF
Figure 4.10
is a force by the hand on the wall (action)Where
is a force by the wall on the hand (reaction)BAFABF
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PHYSICS CHAPTER 4
42
When a book is placed on the table. (refer to Figure 4.11)
If a car is accelerating forward, it is because its tyres are pushing backward on the road and the road is pushing forward on the tyres.
A rocket moves forward as a result of the push exerted on it by the exhaust gases which the rocket has pushed out.
In all cases when two bodies interact, the action and reaction forces act on different bodies.
Figure 4.11
Force by the book on the table (action)
Force by the table on the book (reaction)
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PHYSICS CHAPTER 4
43
Applications of Newtons 2nd law of motion
From the Newtons second law of motion, we arrived at equation
There are five steps in applying the equation above to solve problems in mechanics:
Identify the object whose motion is considered.
Determine the forces exerted on the object.
Draw a free body diagram for each object.
is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it.
Choose a system of coordinates so that calculations may be simplified.
Apply the equation above,
Along xaxis:
Along yaxis:
maFF nett
xxmaF
yy maF
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PHYSICS CHAPTER 4
44
Three wooden blocks connected by a rope of negligible mass are
being dragged by a horizontal force, F in Figure 4.12.
Suppose that F = 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg. Determine
a. the acceleration of blocks system.
b. the tension of the rope, T1 and T2.
Neglect the friction between the floor and the wooden blocks.
Example 4.8:
Figure 4.12
1T
m1 m2 m32T
F
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PHYSICS CHAPTER 4
45
Solution :
a. For the block, m1 = 3 kg
For the block, m2 = 15 kg
For the block, m3 = 30 kg
a
amTFF 11x
(1)
amTTF 221x
(2)
1T
m1
m2
m3
2T
F
aTF 1x 31000
1T a
aTTF 21x 15
2T a
amTF 32x
(3)
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PHYSICS CHAPTER 4
46
Solution :
a. By substituting eq. (3) into eq. (2) thus
Eq. (1)(4) :
b. By substituting the value of acceleration into equations (4) and
(3), therefore
045 aT1 (4)
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PHYSICS CHAPTER 4
47
Two objects of masses m1 = 10 kg and m2 = 15 kg are connected
by a light string which passes over a smooth pulley as shown in
Figure 4.13. Calculate
a. the acceleration of the object of mass 10 kg.
b. the tension in the each string.
(Given g = 9.81 m s2)
Solution :
a. For the object m1= 10 kg,
Example 4.9:
Figure 4.13
m1
m2
1T
gmW 11
amgmTF 111y
(1)agT 1010 a where TTT 21
Simulation 4.2
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PHYSICS CHAPTER 4
48
Solution :
a. For the object m2= 15 kg,
Eq. (1) + (2) :
b. Substitute the value of acceleration into equation (1) thus
Therefore
2T
gmW 22
amTgmF 222y
(2)agT 1515 a
aTgFy 1515
1.96109.8110 T
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PHYSICS CHAPTER 4
49
Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side
and in contact with each another. They are pushed along a smooth
floor under the action of a constant force F of magnitude 200 N applied to A as shown in Figure 4.14. Determine
a. the acceleration of the blocks,
b. the force exerted by A on B.
Solution :
a. Let the acceleration of the blocks is a. Therefore
Example 4.10:
ammF BAx
N 200 kg; 30 kg; 10 Fmm BA
Figure 4.14
A BF
ammF BA
Simulation 4.3
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PHYSICS CHAPTER 4
50
Solution :
b. For the object A,
From the Newtons 3rd law, thus
OR
For the object B,
amFFF ABAx 5.010200 BAFF
a
BAF
A
BABF
a
amFF BABx
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PHYSICS CHAPTER 4
51
1. A block is dragged by forces, F1 and F2 of the magnitude 20 N and 30 N respectively as shown in Figure 4.15. The
frictional force f exerted on the block is 5 N. If the weight of the block is 200 N and it is move horizontally, determine the
acceleration of the block.
(Given g = 9.81 m s2)
ANS. : 1.77 m s2
Exercise 4.3:
50
a
1F
2F
f
20
Figure 4.15
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PHYSICS CHAPTER 4
52
2. One 3.5 kg paint bucket is hanging by a massless cord from
another 3.5 kg paint bucket, also hanging by a massless cord
as shown in Figure 4.16. If the two buckets are pulled upward
with an acceleration of 1.60 m s2 by the upper cord, calculate
the tension in each cord.
(Given g = 9.81 m s2)
ANS. : 39.9 N; 79.8 N
Exercise 4.3:
Figure 4.16
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PHYSICS CHAPTER 4
53
THE ENDNext Chapter
CHAPTER 5 :
Work, Energy and Power
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CHAPTER 5 WORK, ENERGY AND POWER
1
CHAPTER 5:
Work, Energy and Power
(3 Hours)
CHAPTER 5 WORK, ENERGY AND POWER
2
At the end of this chapter, students should be able to:
(a)Define and use work done by a force.
(b) Determine work done from the force
displacement graph.
Learning Outcome:
5.1 Work (1 hour)
sFW
CHAPTER 5 WORK, ENERGY AND POWER
3
5.1 Work, W
Work done by a constant force
is defined as the product of the component of the
force parallel to the displacement times the
displacement of a body.
OR
is defined as the scalar (dot) product between
force and displacement of a body.
CHAPTER 5 WORK, ENERGY AND POWER
4
sFW
FssFW coscos
force of magnitude:F
body theofnt displaceme : s
sF
and between angle the:
Where,
Mathematically :
CHAPTER 5 WORK, ENERGY AND POWER
5
It is a scalar quantity.
Dimension :
The S.I. unit of work is kg m2 s2 or joule (J).
The joule (1 J) is defined as the work done by a force of 1 N
which results in a displacement of 1 m in the direction of
the force.
sFW
22TML W
22 s m kg 1m N 1J 1
CHAPTER 5 WORK, ENERGY AND POWER
6
Work done by a variable force
Figure 5.1 shows a force, F whose magnitude
changes with the displacement, s.
For a small displacement, s1 the force remains almost constant at F1 and work done therefore
becomes W1=F1 s1 .
CHAPTER 5 WORK, ENERGY AND POWER
7
To find the total work done by a variable force, W when the displacement changes from s=s1 to s=s2, we can divide the displacement into N small successive displacements :
s1 , s2 , s3 , , sNThus
FN
F4
s4 sNs1 s2
F/N
s0
F1
s1
W1
NN2211 sFsFsFW ...
Figure 5.1
CHAPTER 5 WORK, ENERGY AND POWER
8
When N , s 0, therefore
2
1
s
sFdsW
graphnt displacemeforce under the area theW
F/N
s/ms1 s20
Work = Area
CHAPTER 5 WORK, ENERGY AND POWER
9
Applications of works equation
Case 1 :
Work done by a horizontal force, F on an object (Figure 4.2).
Case 2 :
Work done by a vertical force, F on an object (Figure 4.3).
0F
s
Figure 5.2
FsW cosFsW
and
90FsW cosJ 0W
andF
s
Figure 5.3
CHAPTER 5 WORK, ENERGY AND POWER
10
Case 3 :
Work done by a horizontal forces, F1 and F2 on an object (Figure 5.4).
Case 4 :
Work done by a force, F and frictional force, f on an object (Figure 5.5).
0cos sFW 11
0cos sFW 22
sFWW nettnett
1F
2F
s
Figure 5.4 sFsFWWW 2121 21nett FFF sFFW 21 and
cos mafFFnett sFW nettnett sfFWnett cos masWnett
f F
Figure 5.5 s
and
OR
CHAPTER 5 WORK, ENERGY AND POWER
11
Caution :
Work done on an object is zero when F = 0 or s = 0 and = 90.
CHAPTER 5 WORK, ENERGY AND POWER
12
Sign for work.
If 0 0 (positive) work done on the system ( by the external force) where energy is transferred to the system.
If 90
CHAPTER 5 WORK, ENERGY AND POWER
13
You push your physics reference book 1.50 m along a horizontal
table with a horizontal force of 5.00 N. The frictional force is 1.60 N.
Calculate
a. the work done by the 5.00 N force,
b. the work done by the frictional force,
c. the total work done on the book.
Solution :
a. Use works equation of constant force,
Example 5.1 :
m 1.50s
N 5.00F
N 1.60f
cosFsWF 0and
Example 5.1 :
CHAPTER 5 WORK, ENERGY AND POWER
14
Solution :
b.
c.
fsW f cos
fF WWW
OR
sFW nett sfFW
180and
180cos1.501.60fW
2.407.50W
1.501.605.00W
CHAPTER 5 WORK, ENERGY AND POWER
15
A box of mass 20 kg moves up a rough plane which is inclined to
the horizontal at 25.0. It is pulled by a horizontal force F of magnitude 250 N. The coefficient of kinetic friction between the box
and the plane is 0.300.
a. If the box travels 3.80 m along the plane, determine
i. the work done on the box by the force F,
ii. the work done on the box by the gravitational force,
iii. the work done on the box by the reaction force,
iv. the work done on the box by the frictional force,
v. the total work done on the box.
b. If the speed of the box is zero at the bottom of the plane,
calculate its speed when it is travelled 3.80 m.
(Given g = 9.81 m s2)
Example 5.2 :
CHAPTER 5 WORK, ENERGY AND POWER
16
Solution :
a. Consider the work done along inclined plane, thus
i.
m 3.800.300; ;N 250 ;kg 20 sFm k
sFW xF cos0and
0cos3.8025cos250FW
25
25
kf
N
F
s
gmW
yF
25cosmg
xF
25sinmg
a
25x
y
CHAPTER 5 WORK, ENERGY AND POWER
17
Solution :
a. ii.
iii.
iv.
smgWg cos25sin 180and 180cos3.8025sin9.8120gW
NsWN cos90and
sfW kf cos180and
180cossNW kf smgFW kf 25cos25sin
3.8025cos9.812025sin2500.300 fW
CHAPTER 5 WORK, ENERGY AND POWER
18
Solution :
a. v.
b. Given
By using equation of work for nett force,
Hence by using the equation of linear motion,
fNgF WWWWW 3230315861W
masW 3.8020223 a
asuv 22 2
0u
3.802.93202v
CHAPTER 5 WORK, ENERGY AND POWER
19
A horizontal force F is applied to a 2.0 kg radiocontrolled car as it moves along a straight track. The force varies with the
displacement of the car as shown in Figure 5.6. Calculate the work
done by the force F when the car moves from 0 to 7 m.
Solution :
Example 5.3 :
5
47
053 6
(N)F
5 (m)s
Figure 5.6
graph under the area sFW
4672
15356
2
1W
CHAPTER 5 WORK, ENERGY AND POWER
20
Exercise 5.1 :
1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless
horizontal table by a constant 16.0 N force directed 25.0 below
the horizontal. Determine the work done on the block by
a. the applied force,
b. the normal force exerted by the table, and
c. the gravitational force.
d. Determine the total work on the block.
(Given g = 9.81 m s2)
ANS. : 31.9 J; (b) & (c) U think; 31.9 J
2. A trolley is rolling across a parking lot of a supermarket. You
apply a constant force to the trolley as it
undergoes a displacement . Calculate
a. the work done on the trolley by the force F,
b. the angle between the force and the displacement of the
trolley.
ANS. : 150 J; 108
N j40i30 F m j3.0i9.0 s
CHAPTER 5 WORK, ENERGY AND POWER
21
Exercise 5.1 :
3.
Figure 5.7 shows an overhead view of three horizontal forces
acting on a cargo that was initially stationary but that now
moves across a frictionless floor. The force magnitudes are
F1 = 3.00 N, F2 = 4.00 N and F3 = 10.0 N. Determine the total work done on the cargo by the three forces during the first
4.00 m of displacement.
ANS. : 15.3 J
3F
1F
2F
y
x
35
50
Figure 5.7
CHAPTER 5 WORK, ENERGY AND POWER
22
At the end of this chapter, students should be able to:
(a) Define and use kinetic energy,
(b) Define and use potential energy:
i. gravitational potential energy,
ii. elastic potential energy for spring,
(c) State and use the principle of conservation of energy.
(d) Explain the workenergy theorem and use the related
equation.
Learning Outcome:
2
2
1mvK
mghU
2
2
1kxU
5.2 Energy And Conservation Of Energy
CHAPTER 5 WORK, ENERGY AND POWER
23
Energy is defined as the systems ability to do work.
The S.I. unit for energy is same to the unit of work (joule, J).
The dimension of energy
is a scalar quantity.
Table 5.1 summarises some common types of energy.
22 TMLWorknergyE
Forms of
EnergyDescription
ChemicalEnergy released when chemical bonds between atoms
and molecules are broken.
Electrical Energy that is associated with the flow of electrical charge.
HeatEnergy that flows from one place to another as a result of
a temperature difference.
InternalTotal of kinetic and potential energy of atoms or molecules
within a body.
CHAPTER 5 WORK, ENERGY AND POWER
24
Forms of
EnergyDescription
Table 5.1
Nuclear Energy released by the splitting of heavy nuclei.
Mass
Energy released when there is a loss of small amount
of mass in a nuclear process. The amount of energy
can be calculated from Einsteins massenergy
equation, E = mc2
Radiant Heat Energy associated with infrared radiation.
SoundEnergy transmitted through the propagation of a series
of compression and rarefaction in solid, liquid or gas.
Mechanicala. Kinetic
b. Gravitational
potentialc. Elastic
potential
Energy associated with the motion of a body.
Energy associated with the position of a body in a
gravitational field.
Energy stored in a compressed or stretched spring.
CHAPTER 5 WORK, ENERGY AND POWER
25
Conservation of energy
5.2.1 Kinetic energy, K is defined as the energy of a body due to its motion.
Equation :
Workkinetic energy theorem
Consider a block with mass, m moving along the horizontal surface (frictionless) under the action of a constant nett force,
Fnett undergoes a displacement, s in Figure 4.8.
2
2
1mvK
body a ofenergy kinetic:K
body a of speed : vbody a of mass : m
where
s
nettF
m
Figure 5.8
maFF nett (1)
CHAPTER 5 WORK, ENERGY AND POWER
26
By using an equation of linear motion:
By substituting equation (2) into (1), we arrive
Therefore
states the work done by the nett force on a body equals the change in the bodys kinetic energy.
as uv 222
s
uva
2
22 (2)
2
22
s
uvmFnett
ifnett KKmumvsF 22
2
1
2
1
KWnett
CHAPTER 5 WORK, ENERGY AND POWER
27
A stationary object of mass 3.0 kg is pulled upwards by a constant
force of magnitude 50 N. Determine the speed of the object when it
is travelled upwards through 4.0 m.
(Given g = 9.81 m s2)
Solution :
The nett force acting on the object is given by
By applying the workkinetic energy theorem, thus
Example 5.4 :
0 m; 4.0 ;N 50; kg 3.0 usFm
F
s
gm
F
gm
9.813.050 mgFFnett
ifnett KKW
02
1 2 mvsFnett
23.02
14.020.6 v
CHAPTER 5 WORK, ENERGY AND POWER
28
A block of mass 2.00 kg slides 0.750 m down an inclined plane that
slopes downward at an angle of 36.9 below the horizontal. If the
block starts from rest, calculate its final speed. You can ignore the
friction. (Given g = 9.81 m s2)
Solution :
Example 5.5 :
s
36.9
0 m; 0.750 ; kg 2.00 usm
N
gm36.9
36.9sinmg36.9cosmg
a
x
y
CHAPTER 5 WORK, ENERGY AND POWER
29
Solution :
Since the motion of the block along the incline surface thus nett
force is given by
By using the workkinetic energy theorem, thus
36.9sinmgFnett
0 m; 0.750 ; kg 2.00 usm
36.9sin9.812.00nettF
ifnett KKW
02
1 2 mvsFnett
22.002
10.75011.8 v
CHAPTER 5 WORK, ENERGY AND POWER
30
An object of mass 2.0 kg moves along the xaxis and is acted on
by a force F. Figure 5.9 shows how F varies with distance
travelled, s. The speed of the object at s = 0 is 10 m s1.
Determine
a. the speed of the object at s = 10 m,
b. the kinetic energy of the object at s = 6.0 m.
Example 5.6 :
10
5
064 10
(N)F
7 (m)s
Figure 5.9
CHAPTER 5 WORK, ENERGY AND POWER
31
Solution :
a.
By using the workkinetic energy theorem, thus
1s m 10 kg; 2.0 um
m 10 tom 0 fromgraph under the area sFW
57106102
11046
2
1W
if KKW
22
2
1
2
1mumvW
22 102.02
12.0
2
132.5 v
CHAPTER 5 WORK, ENERGY AND POWER
32
Solution :
b.
By using the workkinetic energy theorem, thus
m 6 tom 0 fromgraph under the area sFW
10462
1W
if KKW
2
2
1muKW f
2102.02
150 fK
CHAPTER 5 WORK, ENERGY AND POWER
33
Exercise 5.2 :
Use gravitational acceleration, g = 9.81 m s2
1. A bullet of mass 15 g moves horizontally at velocity of
250 m s1.It strikes a wooden block of mass 400 g placed at rest
on a floor. After striking the block, the bullet is embedded in the
block. The block then moves through 15 m and stops. Calculate
the coefficient of kinetic friction between the block and the floor.
ANS. : 0.278
2. A parcel is launched at an initial speed of 3.0 m s1 up a rough
plane inclined at an angle of 35 above the horizontal. The
coefficient of kinetic friction between the parcel and the plane is
0.30. Determine
a. the maximum distance travelled by the parcel up the plane,
b. the speed of the parcel when it slides back to the starting
point.
ANS. : 0.560 m; 1.90 m s1
CHAPTER 5 WORK, ENERGY AND POWER
34
5.2.2 Potential Energy
is defined as the energy stored in a body or system because
of its position, shape and state.
Gravitational potential energy, U
is defined as the energy stored in a body or system because
of its position.
Equation :
The gravitational potential energy depends only on the height
of the object above the surface of the Earth.
mghU
energy potential nalgravitatio : U
position initial thefrombody a ofheight : h
where
body a of mass : mgravity todueon accelerati : g
CHAPTER 5 WORK, ENERGY AND POWER
35
Workgravitational potential energy theorem
Consider a book with mass, m is dropped from height, h1 to
height, h2 as shown in the Figure 5.10.
states the change in gravitational potential energy as the negative of the work done by the gravitational force.
1h
gm
gm
2h
s
Figure 5.10
21g hhmgmgsW
The work done by the gravitational force
(weight) is
fig UUmghmghW 21 UUUW ifg
UW
Therefore in general,
CHAPTER 5 WORK, ENERGY AND POWER
36
Negative sign in the equation indicates that
When the body moves down, h decreases, the
gravitational force does positive work because U 0.
For calculation, use
if UUUW
energy potential nalgravitatio final : fUwhere
force nalgravitatio aby done work : W
energy potential nalgravitatio initial : iU
CHAPTER 5 WORK, ENERGY AND POWER
37
In a smooth pulley system, a force F is required to bring an object of mass 5.00 kg to the height of 20.0 m at a constant
speed of 3.00 m s1 as shown in Figure 5.11. Determine
a. the force, F
b. the work done by the force, F.
(Given g = 9.81 m s2)
Example 5.7 :
Figure 5.11
F
m 20.0
CHAPTER 5 WORK, ENERGY AND POWER
38
Solution :
a. Since the object moves at the constant
speed, thus
b. From the equation of work,
1s m 3.00constant m; 20.0 kg; 5.00 vhsm
0nettFmgF
F
s
gm
F
gm
Constant
speedFsW cos 0and
OR
FsW cosmghUW
0and
CHAPTER 5 WORK, ENERGY AND POWER
39
Elastic potential energy, Us is defined as the energy stored in in elastic materials as the
result of their stretching or compressing.
Springs are a special instance of device which can store
elastic potential energy due to its compression or
stretching.
Hookes Law states the restoring force, Fs of spring is directly proportional to the amount of stretch or
compression (extension or elongation), x if the limit of proportionality is not exceeded
OR xFs
kxFs
spring of force restoring the: sF
)(n compressioor stretch ofamount the: if xxx
constant forceor constant spring the:k
where
CHAPTER 5 WORK, ENERGY AND POWER
40
Negative sign in the equation indicates that the direction of Fsis always opposite to the direction of the amount of stretch or
compression (extension), x.
Case 1:
The spring is hung vertically and its is stretched by a suspended
object with mass, m as shown in Figure 5.12.
The spring is in equilibrium, thus
Initial position
Final position
sF
gmW
x
Figure 5.12
mgWFs
CHAPTER 5 WORK, ENERGY AND POWER
41Figure 5.13
(Equilibrium position)
Case 2:
The spring is attached to an object and it is stretched and
compre5sed by a force, F as shown in Figure 5.13.
sF
F
0x
0x
sF
F
x
x
negative is sFpositive is x
positive is sFnegative is x
0 sF0x
The spring is in equilibrium,
hence
FFs
CHAPTER 5 WORK, ENERGY AND POWER
42
Caution:
For calculation, use :
Dimension of spring constant, k :
The unit of k is kg s2 or N m1
From the Hookes law (without sign), a restoring force, Fsagainst extension of the spring, x graph is shown in Figure 5.14.
FkxFs
2s MTx
Fk
force applied : Fwhere
F
sF
0 x1x
graph under the area xFW s
1FxW2
1 11 xkxW
2
1
s21 UkxW
2
1
Figure 5.14
CHAPTER 5 WORK, ENERGY AND POWER
43
The equation of elastic potential energy, Us for compressing or stretching a spring is
The workelastic potential energy theorem,
Notes :
Workenergy theorem states the work done by the nett
force on a body equals the change in the bodys total energy
OR
xF2
1kx
2
1U s
2s
ifnett EEEW
sUW 2
i
2
fsisf kx2
1kx
2
1UUW OR
CHAPTER 5 WORK, ENERGY AND POWER
44
A force of magnitude 800 N caused an extension of 20 cm on a
spring. Determine the elastic potential energy of the spring when
a. the extension of the spring is 30 cm.
b. a mass of 60 kg is suspended vertically from the spring.
(Given g = 9.81 m s2)
Solution :
From the Hookes law,
a. Given x=0.300 m,
Example 5.8 :
m 0.200 N; 800 xF
kxFFs 0.20800 k
2
2
1kxU s
23 0.3001042
1sU
CHAPTER 5 WORK, ENERGY AND POWER
45
Solution :
b. Given m=60 kg. When the spring in equilibrium, thus
Therefore
0nettFmgFs mgkx
9.8160104 3 x
2
2
1kxU s
23 0.1471042
1sU
sF
gmW
x
CHAPTER 5 WORK, ENERGY AND POWER
46
5.2.3 Principle of conservation of energy
states in an isolated (closed) system, the total energy of that system is constant.
According to the principle of conservation of energy, we get
The initial of total energy = the final of total energy
Conservation of mechanical energy
In an isolated system, the mechanical energy of a system is the
sum of its potential energy, U and the kinetic energy, K of theobjects are constant.
OR
fi EE
constant UKE
OR
ffii UKUK
CHAPTER 5 WORK, ENERGY AND POWER
47
Before After
cm 30
x
Figure 5.15
A 1.5 kg sphere is dropped from a height of
30 cm onto a spring of spring constant,
k = 2000 N m1 . After the block hits the spring, the spring experiences maximum
compression, x as shown in Figure 5.15.
a. Describe the energy conversion
occurred after the sphere is
dropped onto the spring until the
spring experiences maximum
compression, x.
b. Calculate the speed of the sphere just
before strikes the spring.
c. Determine the maximum compression, x.
(Given g = 9.81 m s2)
Example 5.9 :
CHAPTER 5 WORK, ENERGY AND POWER
48
The spring is not stretched
hence Us = 0. The sphere is
at height h1 above ground with speed, v just before strikes the spring. Therefore
The sphere is at height h2above the ground after
compressing the spring by x. The speed of the sphere at
this moment is zero. Hence
The spring is not stretched
hence Us = 0. The sphere is
at height h0 above ground
therefore U = mgh0 and it is
stationary hence K = 0.
(2)
v
1h
(3)
x
2h
cm 30h
0h
(1)
01 mghE 212 mv2
1mghE 223 kx
2
1mghE
Solution :
a.
CHAPTER 5 WORK, ENERGY AND POWER
49
Solution :
b. Applying the principle of conservation of energy involving the
situation (1) and (2),
210 mvhhmg2
1
21 EE2
10 mvmghmgh2
1
0.309.812 v
1m N 2000 m; 0.30 kg; 1.5 khm
and 10 hhh
ghv 2
CHAPTER 5 WORK, ENERGY AND POWER
50
Solution :
c. Applying the principle of conservation of energy involving the
situation (2) and (3),
2221 kxmvhhmg2
1
2
1
32 EE2
22
1 kxmghmvmgh2
1
2
1
1m N 2000 m; 0.30 kg; 1.5 khm
and 21 hhx
22 20002
12.431.5
2
19.811.5 xx
CHAPTER 5 WORK, ENERGY AND POWER
51
A bullet of mass, m1=5.00 g is fired into a wooden block of mass,
m2=1.00 kg suspended from some light wires as shown in Figure 5.16. The block, initially at rest. The bullet embeds in the block, and
together swing through a height, h=5.50 cm. Calculate
a. the initial speed of the bullet.
b. the amount of energy lost to the surrounding.
(Given g = 9.81 m s2)
Example 5.10 :
Figure 5.16
1m 2m
21 mm
h1u
CHAPTER 5 WORK, ENERGY AND POWER
52
(1)
1m 2m1u
02u
(3)
h
21 mm
012v
(2)
21 mm 12u
m 105.50 kg; 1.00kg; 105.00 23 hmm 21
32 EE
ghmmumm 211221 2
2
1
2105.509.8122 ghu12
UK
Solution :
a.
Applying the principle of conservation of energy involving the
situation (2) and (3),
CHAPTER 5 WORK, ENERGY AND POWER
53
Solution :
Applying the principle of conservation of linear momentum
involving the situation (1) and (2),
b. The energy lost to the surrounding, Q is given by
m 105.50 kg; 1.00kg; 105.00 23 hmm 21
21 pp
122111 ummum 1.041.00105.00105.00 33 1u
21 EEQ
22
1
2
11221
211 ummumQ
2323 1.041.00105.002
1209105.00
2
1 Q
CHAPTER 5 WORK, ENERGY AND POWER
54
Objects P and Q of masses 2.0 kg and 4.0 kg respectively are
connected by a light string and suspended as shown in Figure
5.17. Object Q is released from rest. Calculate the speed of Q at
the instant just before it strikes the floor.
(Given g = 9.81 m s2)
Example 5.11 :
Figure 5.17
P
Q
m 2
Smooth
pulley
CHAPTER 5 WORK, ENERGY AND POWER
55
Solution :
Applying the principle of conservation of mechanical energy,
0 m; 2 kg; 4.0kg; 2.0 QP uhmm
fi EE2
Q2
PPQ2
1
2
1vmvmghmghm
QPPQ KKUU
Initial
P
Q
m 2
Smooth
pulley
P
Qm 2
Smooth
pulley
v
v
Final
22 4.02
12.0
2
129.812.029.814.0 vv
CHAPTER 5 WORK, ENERGY AND POWER
56
Exercise 5.3 :
Use gravitational acceleration, g = 9.81 m s2
1. If it takes 4.00 J of work to stretch a spring 10.0 cm from its
initial length, determine the extra work required to stretch it an
additional 10.0 cm.
ANS. : 12.0 J
2. A book of mass 0.250 kg is placed on top of a light vertical
spring of force constant 5000 N m1 that is compressed by 10.0
cm. If the spring is released, calculate the height of the book rise
from its initial position.
ANS. : 10.2 m
3. A 60 kg bungee jumper jumps from a bridge. She is tied to a
bungee cord that is 12 m long when unstretched and falls a total
distance of 31 m. Calculate
a. the spring constant of the bungee cord.
b. the maximum acceleration experienced by the jumper.
ANS. : 100 N m1; 22 m s2
CHAPTER 5 WORK, ENERGY AND POWER
57
Exercise 5.3 :
4.
A 2.00 kg block is pushed against a light spring of the force
constant, k = 400 N m1, compressing it x =0.220 m. When the block is released, it moves along a frictionless horizontal surface
and then up a frictionless incline plane with slope =37.0 as shown in Figure 5.18. Calculate
a. the speed of the block as it slides along the horizontal
surface after leaves the spring.
b. the distance travelled by the block up the incline plane before
it slides back down.
ANS. : 3.11 m s1; 0.81 m
Figure 5.18
CHAPTER 5 WORK, ENERGY AND POWER
58
Exercise 5.3 :
5.
A ball of mass 0.50 kg is at point A with initial speed, u =4 m s1
at a height of 10 m as shown in Figure 5.19 (Ignore the frictional
force). Determine
a. the total energy at point A,
b. the speed of the ball at point B where the height is 3 m,
c. the speed of the ball at point D,
d. the maximum height of point C so that the ball can pass over
it.
ANS. : 53.1 J; 12.4 m s1; 14.6 m s1; 10.8 m
u
m 10
A
B
C
DFigure 5.19
CHAPTER 5 WORK, ENERGY AND POWER
59
At the end of this chapter, students should be able to:
(a) Define and use power:
Average power,
Instantaneous Power,
(b) Derive and apply the formulae
(c) Define and use mechanical efficiency,
and the consequences of heat dissipation.
Learning Outcome:
5.3 Power and mechanical efficiency (1 hour)
dt
dWP
vFP
t
WPav
100%input
output
P
P
CHAPTER 5 WORK, ENERGY AND POWER
60
5.3 Power and mechanical efficiency
5.3.1 Power, P is defined as the rate at which work is done.
OR the rate at which energy is transferred.
If an amount of work, W is done in an amount of time t by aforce, the average power, Pav due to force during that timeinterval is
The instantaneous power, P is defined as the instantaneousrate of doing work, which can be write as
t
E
t
WPav
dt
dW
t
WP
0tlimit
CHAPTER 5 WORK, ENERGY AND POWER
61
is a scalar quantity.
The dimension of the power is
The S.I. unit of the power is kg m2 s3 or J s1 or watt (W).
Unit conversion of watt (W), horsepower (hp) and foot pounds
per second (ft. lb s1)
Consider an object that is moving at a constant velocity v along a frictionless horizontal surface and is acted by a constant force,
F directed at angle above the horizontal as shown in Figure 5.20. The object undergoes a displacement of ds.
3222
TMLT
TML
t
WP
1s lb ft. 550 W746hp 1
Figure 5.20
F
sd
CHAPTER 5 WORK, ENERGY AND POWER
62
Therefore the instantaneous power, P is given by
OR
dt
dWP
vFP
dsFdW cos
FvP cos
and
dt
dsFP
cos
dt
dsv and
where
vF
and between angle the:
force of magnitude:F velocityof magnitude : v
CHAPTER 5 WORK, ENERGY AND POWER
63
An elevator has a mass of 1.5 Mg and is carrying 15 passengers
through a height of 20 m from the ground. If the time taken to lift
the elevator to that height is 55 s. Calculate the average power
required by the motor if no energy is lost. (Use g = 9.81 m s2 and the average mass per passenger is 55 kg)
Solution :
M = mass of the elevator + mass of the 15 passengers
M = 1500 + (5515) = 2325 kg
According to the definition of average power,
Example 5.12 :
t
MghPav
t
EPav
s 55 m; 20 th
CHAPTER 5 WORK, ENERGY AND POWER
64
An object of mass 2.0 kg moves at a constant speed of 5.0 m s1
up a plane inclined at 30 to the horizontal. The constant frictional
force acting on the object is 4.0 N. Determine
a. the rate of work done against the gravitational force,
b. the rate of work done against the frictional force,
c. the power supplied to the object. (Given g = 9.81 m s2 )
Solution :
Example 5.13 :
N 4.0 constant;s m 5.0 kg; 2.0 1 fvm
30
f
N
gmW
30cosmg
30sinmg
v
30x
y
s
CHAPTER 5 WORK, ENERGY AND POWER
65
Solution :
a. the rate of work done against the gravitational force is given by
t
smg
t
Wg cos30sin
N 4.0 constant;s m 5.0 kg; 2.0 1 fvm
180and
t
smg
t
Wg 30sin
t
sv and
vmgt
Wg 30sin
5.030sin9.812.0
t
Wg
OR vFt
Wg
gcos
180cos30sin vmgt
Wg
CHAPTER 5 WORK, ENERGY AND POWER
66
Solution :
b. The rate of work done against the frictional force is
c. The power supplied to the object, Psupplied
= the power lost against gravitational and frictional forces, Plost
N 4.0 constant;s m 5.0 kg; 2.0 1 fvm
180andfvt
W fcos
t
W
t
WP
fg
supplied
CHAPTER 5 WORK, ENERGY AND POWER
67
5.3.2 Mechanical efficiency, Efficiency is a measure of the performance of a machines,
engine and etc...
The efficiency of a machine is defined as the ratio of the useful
(output) work done to the energy input.
is a dimensionless quantity (no unit).
Equations:
100% in
out
E
W
OR
100% in
out
P
P
where system by the producedpower :outP
system a tosuppliedpower : inP
CHAPTER 5 WORK, ENERGY AND POWER
68
Notes :
In practice, Pout< Pin hence < 100%. The system loses energy to its surrounding because it may
have encountered resistances such as surface friction or
air resistance.
The energy which is dissipated to the surroundings, may
be in the form of heat or sound.
A 1.0 kW motor is used to lift an object of mass 10 kg vertically
upwards at a constant speed. The efficiency of the motor is 75 %.
Determine
a. the rate of heat dissipated to the surrounding.
b. the vertical distance travelled by the object in 5.0 s.
(Given g = 9.81 m s2 )
Example 5.14 :
CHAPTER 5 WORK, ENERGY AND POWER
69
Solution :
a. The output power of the motor is given by
Therefore the rate of heat dissipated to the surrounding is
b.
Since the speed is constant hence the vertical distance in 5.0 s
is
W1000 75%; kg; 10.0 inPm
%100in
out
P
P
1001000
75 outP
7501000dissipatedheat of Rate outin PP
FvPout cos0where and mgF
0cosmgvPout
t
hv
CHAPTER 5 WORK, ENERGY AND POWER
70
Exercise 5.4 :
Use gravitational acceleration, g = 9.81 m s2
1. A person of mass 50 kg runs 200 m up a straight road inclined
at an angle of 20 in 50 s. Neglect friction and air resistance.
Determine
a. the work done,
b. the average power of the person.
ANS. : 3.36104 J; 672 W
2. Electrical power of 2.0 kW is delivered to a motor, which has an
efficiency of 85 %. The motor is used to lift a block of mass
80 kg. Calculate
a. the power produced by the motor.
b. the constant speed at which the block being lifted vertically
upwards by the force produced by the motor.
(neglect air resistance)
ANS. : 1.7 kW; 2.17 m s1
CHAPTER 5 WORK, ENERGY AND POWER
71
Exercise 5.4 :
3.
A car of mass 1500 kg moves at a constant speed v up a road
with an inclination of 1 in 10 as shown in Figure 5.21. All
resistances against the motion of the car can be neglected. If
the engine car supplies a power of 12.5 kW, calculate the
speed v.
ANS. : 8.50 m s1
Figure 5.21
10 1
PHYSICS CHAPTER 6
1
CHAPTER 6:
Circular motion
(3 Hours)
PHYSICS CHAPTER 6
2
At the end of this chapter, students should be able to:
Describe graphically the uniform circular motion.
In terms of velocity with constant magnitude (only the
direction of the velocity changes).
Learning Outcome:
6.1 Uniform circular motion (1 hour)
PHYSICS CHAPTER 6
3
6.1 Uniform circular motion is defined as a motion in a circle (circular arc) at a constant
speed.
Consider an object which does move with uniform circular
motion as shown in Figure 6.1.
Figure 6.1
r
O
s
The length of a circular arc, s is given by
rs
pathcircular theof radius : r
subtends arc which theangle : where
radianin circle theof centre theto
PHYSICS CHAPTER 6
4
It is directed tangentially to the circular path and always
perpendicular to the radius of the circular path as shown in
Figure 6.2.
In uniform circular motion, the magnitude of the linear velocity
(speed) of an object is constant but the direction is
continually changing.
The unit of the tangential (linear) velocity is m s1.
6.1.1 Linear (tangential) velocity ,
r
O
v
r
v
r
vFigure 6.2
v
PHYSICS CHAPTER 6
5
The linear velocity, v is difficult to measure but we can measure the period, T of an object in circular motion.
Period, T
is defined as the time taken for one complete revolution (cycle/rotation).
The unit of the period is second (s).
Frequency, f
is defined as the number of revolutions (cycles/rotations) completed in one second.
The unit of the frequency is hertz (Hz) or s1.
Equation :
Let the object makes one complete revolution in circular motion, thus
the distance travelled is (circumference of the circle),
the time interval is one period, T.
Tf
1
r2
PHYSICS CHAPTER 6
6
From the definition of speed,
If therefore
Note:
The unit of angular velocity (angular frequency) is rad s1
(radian per second).
Unit conversion of angle, :
interval time
distance of changev
T
rv
2OR rfv 2
fT
22
rv
pathcircular theof radius : rfrequency)(angular locity angular ve :
where
360rad 2
180rad
PHYSICS CHAPTER 6
7
At the end of this chapter, students should be able to:
Define and use centripetal acceleration and use
centripetal acceleration,
Define and solve problem on centripetal force,
Learning Outcome:
6.2 Centripetal force (2 hours)
r
va
2
c
r
mvF
2
c
PHYSICS CHAPTER 6
8
Figure 6.3 shows a particle moving with constant speed in a
circular path of radius, r with centre at O. The particle moves
from A to B in a time, t.
6.2.1 Centripetal (radial) acceleration,rc aa
or
Figure 6.3
1v
2v
The arc length AB is given by
The velocities of the particle at A
and B are v1 and v2 respectively where
rs
r
s
vvv 21
(1)
PHYSICS CHAPTER 6
9
Let PQ and PR represent the velocity vectors v1 and v2respectively, as shown in Figure 6.4.
Then QR represent the change in velocity vector v of the particle in time interval t. Since the angle between PQ and PR is small hence
By equating (1) and (2) then
12 vvv
2v
1v
P Q
RFigure 6.4
PQQR vv
v
v (2)
v
v
r
s
PHYSICS CHAPTER 6
10
Dividing by time, t, thus
t
v
v
1
t
s
r
1
v
a
r
v
r
va
2
c
pathcircular of radius : r
onaccelerati lcentripeta : cawhere
velocity gential)linear(tan : v
OR vra 2c
frequency)(angular locity angular ve :
PHYSICS CHAPTER 6
11
ca
ca
ca
caca
ca
Figure 6.5
The centripetal acceleration is defined as the acceleration of
an object moving in circular path whose direction is
towards the centre of the circular path and whose
magnitude is equal to the square of the speed divided by
the radius.
The direction of centripetal (radial) acceleration is always
directed toward the centre of the circle and perpendicular to
the linear (tangential) velocity as shown in Figure 6.5.
PHYSICS CHAPTER 6
12
For uniform circular motion, the magnitude of the centripetal
acceleration always constant but its direction continuously
changes as the object moves around the circular path.
Because of
therefore we can obtain the alternative expression of centripetal
acceleration is
2
2
cT
ra
4
T
rv
2
r
a
2
Tr
c
2
PHYSICS CHAPTER 6
13
A motorbike moving at a constant speed 20.0 m s1 in a circular
track of radius 25.0 m. Calculate
a. the centripetal acceleration of the motorbike,
b. the time taken for the motorbike to complete one revolution.
Solution :
a. From the definition of the centripetal acceleration, thus
b. From the alternate formula of the centripetal acceleration, hence
Example 6.1 :
m 25.0 ;s m 20.0 1 rv
r
va
2
c
2
2
cT
ra
4
25.0
20.02
ca
2
25.0416.0
T
2
T
rv
2OR
PHYSICS CHAPTER 6
14
A car initially travelling eastward turns north by travelling in a
circular path at uniform speed as shown in Figure 6.6. The length
of the arc ABC is 235 m and the car completes the turn in 36.0 s.
Determine
a. the acceleration when the car is at B located at an angle of
35.0,
b. the cars speed,
c. its average acceleration during the 36.0 s interval.
Example 6.2 :
Figure 6.6
PHYSICS CHAPTER 6
15
Solution :
a. The period of the car is given by
The radius of the circular path is
Therefore the magnitude of the centripetal acceleration is
s 36.0 m, 235 tsABC
36.044 tTs 144T
rsABC
2
235 r
2
2
cT
ra
4 2
2
144
1504ca
PHYSICS CHAPTER 6
16
Solution :
b. From the definition of the speed, thus
c. 1st method :
By using the triangle method for vector addition, thus the change
in the velocity is given by
s 36.0 m, 235 tsABC
t
s
t
sv ABC
36.0
235v
Cv
Av
AC vvv
2A2
C vvv
22 6.536.53 v
PHYSICS CHAPTER 6
17
Solution :
Therefore the magnitude of the average acceleration is
and its direction :
s 36.0 m, 235 tsABC
t
vaav
36.0
9.24ava
A
C1
v
v tan
6.53
6.53tan 1
PHYSICS CHAPTER 6
18
Solution :
c. 2nd method :
xcomponent :
ycomponent :
s 36.0 m, 235 tsABC
t
vv
t
va AxCxx
xav
36.0
6.530
xava
t
vv
t
va
AyCyy
yav
36.0
06.53
yava
PHYSICS CHAPTER 6
19
Solution :
Therefore the magnitude of the average acceleration is
and
s 36.0 m, 235 tsABC
2yav
2
xavavaaa
22 0.1810.181 ava
xav
yav1
a
a tan
0.181
0.181tan 1
PHYSICS CHAPTER 6
20
A boy whirls a marble in a horizontal circle of radius 2.00 m and at
height 1.65 m above the ground. The string breaks and the marble
flies off horizontally and strikes the ground after traveling a
horizontal distance of 13.0 m. Calculate
a. the speed of the marble in the circular path,
b. the centripetal acceleration of the marble while in the circular
motion.
(Given g = 9.81 m s2)
Solution :
Example 6.3 :
1.65 m
Before
13.0 m
u
After
u
r =2.00 m
1.65 m
PHYSICS CHAPTER 6
21
Solution :
a. From the diagram :
The time taken for the marble to strike the ground is
The initial speed of the marble after the string breaks is equal to
the tangential speed of the marble in the horizontal circle.
Therefore
0 ; yx uuu
29.812
101.65 t
2
2
1gttus yy
m 1.65 ; m 13.0 yx ss
0.58013.0 utus xx
PHYSICS CHAPTER 6
22
Solution :
b. From the definition of the centripetal acceleration, thus
r
u
r
vac
22
2.00
22.42
ca
PHYSICS CHAPTER 6
25
6.3 Centripetal force6.3.1 Equation of centripetal force
From Newtons second law of motion, a force must be associated with the centripetal acceleration. This force is
known as the centripetal force and is given by
amFF nett
cc amF
mvmrr
mvF 2
2
c
caa
vrr
va 2
2
c
where cFF
and
and
force lcentripeta :cFwhere
PHYSICS CHAPTER 6
26
ca
cF
cF
cF
ca
ca
v
v
v
The centripetal force is defined as a force acting on a body
causing it to move in a circular path of magnitude
and its always directed towards the centre of the circular
path.
Its direction is in the same direction of the centripetal
acceleration as shown in Figure 6.8.
Figure 6.8
r
mvF
2
c
PHYSICS CHAPTER 6
27
PHYSICS CHAPTER 6
28
cF
ca v
cF
cF
ca
ca
v
v
v
v
0Fc
0Fc
0ac
0ac
If the centripetal force suddenly stops to act on a body in the
circular motion, the body flies off in a straight line with the
constant tangential (linear) speed as show in Figure 6.9.
Note :
In uniform circular motion, the nett force on the system is
centripetal force.
The work done by the centripetal force is zero but the
kinetic energy of the body is not zero and given by
Figure 6.9
222 mr2
1mv
2
1K
Simulation 6.1
PHYSICS CHAPTER 6
29
As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.
As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.
As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion
PHYSICS CHAPTER 6
30
Without a centripetal force, an
object in motion continues along a
straightline path.
With a centripetal force, an object in
motion will be accelerated and change its
direction.
PHYSICS CHAPTER 6
31
Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient.
PHYSICS CHAPTER 6
32
PHYSICS CHAPTER 6
33
Conical Pendulum
Example 6.4 :
Figure 6.10 shows a conical pendulum
with a bob of mass 80.0 kg on a 10.0 m
long string making an angle of 5.00 to the
vertical.
a. Sketch a free body diagram of the bob.
b. Determine
i. the tension in the string,
ii. the speed and the period of the bob,
iii. the radial acceleration of the bob.
(Given g =9.81 m s2)
6.3.2 Examples of uniform circular motion
Figure 6.10
PHYSICS CHAPTER 6
34
Solution :
a. The free body diagram of the bob :
b. i. From the diagram,
5.00 ;m 10.0 ;kg 80.0 lm
gm
T
T cos
T sin
0 yFmgT cos
ca
PHYSICS CHAPTER 6
35
The centripetal force is contributed
by the horizontal component of the
tension.
Solution :
b. ii.
5.00 ;m 10.0 ;kg 80.0 lm
cx FF
r
mvT
2
sin
l
mvT
2
sinsin
r
ll
r sin
lr sin
m
Tlv
2sin
80.0
5.00sin10.07882
v
PHYSICS CHAPTER 6
36
Solution :
b. ii. and the period of the bob is given by
iii. From the definition of the radial acceleration, hence
5.00 ;m 10.0 ;kg 80.0 lm
T
rv
2
T
lv
sin2
T
5.00sin10.020.865
l
va
2
rsin
5.00sin10.0
0.8652
ra
r
va
2
r
PHYSICS CHAPTER 6
37
Centre of
circle
Motion rounds a curve on a flat (unbanked) track (for car,
motorcycle, bicycle, etc)
Example 6.5 :
A car of mass 2000 kg rounds a circular turn of radius 20 m. The
road is flat and the coefficient of friction between tires and the road
is 0.70.
a. Sketch a free body diagram of the car.
b. Determine the maximum cars speed without skidding.
(Given g = 9.81 m s2)
Solution :
a. The free body diagram of the car :
gm
N
f
0.70 ;m 20 ;kg 2000 rm
ca
Picture 6.1
PHYSICS CHAPTER 6
38
Solution :
b. From the diagram in (a),
ycomponent :
xcomponent : The centripetal force is provided by the frictional
force between the wheel (4 tyres) and the road.
Therefore
0.70 ;m 20 ;kg 2000 rm
0yF mgN
r
mvF
2
x
r
mvf
2
rgv r
mvmg
2
PHYSICS CHAPTER 6
39
T
gm
r
ca
Motion in a horizontal circle
Example 6.6 :
A ball of mass 150 g is attached to one end of a string 1.10 m long.
The ball makes 2.00 revolution per second in a horizontal circle.
a. Sketch the free body diagram for the ball.
b. Determine
i. the centripetal acceleration of the ball,
ii. the magnitude of the tension in the string.
Solution :
a. The free body diagram for the ball :
Hz 2.00 ;m 1.10 ;kg 0.150 frlm
PHYSICS CHAPTER 6
40
Solution :
b. i. The linear speed of the ball is given by
Therefore the centripetal acceleration is
ii. From the diagram in (a), the centripetal force enables the ball
to move in a circle is provided by the tension in the string.
Hence
Hz 2.00 ;m 1.10 ;kg 0.150 frlm
rfT
rv
2
2
2.001.102v
r
va
2
c 1.10
13.82
ca
ccx maFF cmaT
PHYSICS CHAPTER 6
41
Motion in a vertical circle
Example 6.7 :
A small remote control car with mass 1.20 kg moves at a constant
speed of v = 15.0 m s1 in a vertical circle track of radius 3.00 m as shown in Figure 6.12. Determine the magnitude of the reaction
force exerted on the car by the track at
a. point A,
b. point B.
(Given g = 9.81 m s2)
m 3.00
v
v
A
B
Figure 6.12
PHYSICS CHAPTER 6
42
Solution :
a. The free body diagram of the car at point A :
1s m 15.0 ;m 3.00 ;kg 1.20 vrm
gm
AN
ca
r
mvF
2
r
mvmgN
2
A
3.00
15.01.209.811.20
2
AN
PHYSICS CHAPTER 6
43
Solution :
b. The free body diagram of the car at point B :
1s m 15.0 ;m 3.00 ;kg 1.20 vrm
BN
ca
r
mvF
2
r
mvmgN
2
B
3.00
15.01.209.811.20
2
BN
gm
PHYSICS CHAPTER 6
44
A rider on a Ferris wheel moves in a vertical circle of radius,
r = 8 m at constant speed, v as shown in Figure 6.13. If the time taken to makes one rotation is 10 s and the mass of the rider is
60 kg, Calculate the normal force exerted on the rider
a. at the top of the circle,
b. at the bottom of the circle.
(Given g = 9.81 m s2)
Example 6.8 :
v
v
Figure 6.13
PHYSICS CHAPTER 6
45
Solution :
a. The constant speed of the rider is
The free body diagram of the rider at the top of the circle :
s 10 ;m 8 ;kg 60 Trm
T
rv
2 10
82v
ca
gm
tN
r
mvF
2
r
mvNmg
2
t
8
5.03609.8160
2
tN
PHYSICS CHAPTER 6
46
Solution :
b. The free body diagram of the rider at the bottom of the circle :
s 10 ;m 8 ;kg 60 Trm
ca
gm
r
mvF
2
r
mvmgN
2
b
8
5.03609.8160
2
bNbN
PHYSICS CHAPTER 6
47
A sphere of mass 5.0 kg is tied to an inelastic string. It moves in a
vertical circle of radius 55 cm at a constant speed of 3.0 m s1 as
shown in Figure 6.14. By the aid of the free body diagram,
determine the tension in the string at points A, D and E.
(Given g = 9.81 m s2)
Example 6.9 :
Figure 6.14
A
D
E 3.0 m s1
3.0 m s1
3.0 m s1
PHYSICS CHAPTER 6
48
Solution :
The free body diagram of the sphere at :
Point A,
Point D,
1s m 0.3 ;m 55.0 ;kg 0.5 vrm
ca
r
mvF
2
r
mvmgTA
2
0.55
3.05.09.815.0
2
AT
A
gmAT
ca
D
gm
DT
r
mvTD
2
0.55
3.05.02
DT
PHYSICS CHAPTER 6
49
Solution :
The free body diagram of the sphere at :
Point E,
Caution :
For vertical uniform circular motion only,
the normal force or tension is maximum at the bottom of
the circle.
the normal force or tension is minimum at the top of the
circle.
1s m 0.3 ;m 55.0 ;kg 0.5 vrm
ca r
mvmgTE
2
0.55
3.05.09.815.0
2
ETE
gm
ET
PHYSICS CHAPTER 6
50
Exercise 6.2 :
Use gravitational acceleration, g = 9.81 m s2
1. A cyclist goes around a curve of 50 m radius at a speed of
15 m s1. The road is banked at an angle to the horizontal and the cyclist travels at the right angle with the surface of the road.
The mass of the bicycle and the cyclist together equals 95 kg.
Calculate
a. the magnitude of the centripetal acceleration of the cyclist,
b. the magnitude of the normal force which the road exerts on
the bicycle and the cyclist,
c. the angle .ANS. : 4.5 m s2; 1.02 kN; 24.6
PHYSICS CHAPTER 6
51
Exercise 6.2 :
2. A ball of mass 0.35 kg is attached to the end of a horizontal
cord and is rotated in a circle of radius 1.0 m on a frictionless
horizontal surface. If the cord will break when the tension in it
exceeds 80 N, determine
a. the maximum speed of the ball,
b. the minimum period of the ball.
ANS. : 15.1 m s1; 0.416 s
Figure 6.14
3. A small mass, m is set on the surface of a sphere as shown in Figure 6.14.
If the coefficient of static friction is s= 0.60, calculate the angle would the mass start sliding.
ANS. : 31
m
O
PHYSICS CHAPTER 6
52
Exercise 6.2 :
4. A ball of mass 1.34 kg is connected
by means of two massless string to
a vertical rotating rod as shown in
Figure 6.15. The strings are tied to
the rod and are taut. The tension in
the upper string is 35 N.
a. Sketch a free body diagram for
the ball.
b. Calculate
i. the magnitude of the tension
in the lower string,
ii. the nett force on the ball,
iii. the speed of the ball.
ANS. : 8.74 N; 37.9 N (radially
inward); 6.45 m s1
Figure 6.15
PHYSICS CHAPTER 6
53
THE END
Next ChapterCHAPTER 7 :
Gravitation
PHYSICS CHAPTER 7
1
CHAPTER 7:
Gravitation
(2 Hours)
ww
w.k
ms.m
atr
ik.e
du
.my/p
hysic
s
PHYSICS CHAPTER 7
In this chapter, we learns about
7.1 Gravitational force and field strength
7.2 Gravitational potential
7.3 Satellite motion in a circular orbit
2
PHYSICS CHAPTER 7
7.1 Gravitational Force and Field
Strength
7.1.1 Newtons law of gravitation
7.1.2 Gravitational Field
7.1.3 Gravitational force and field strength
3
PHYSICS CHAPTER 7
4
At the end of this chapter, students should be able to:
State and use the Newtons law of gravitation,
Learning Outcome:
7.1 Newtons law of gravitation (1 hour)
2
21
r
mmGFg
ww
w.k
ms.m
atr
ik.e
du
.my/p
hysic
s
PHYSICS CHAPTER 7
5
7.1.1 Newtons law of gravitation
States that a magnitude of an attractive force between two
point masses is directly proportional to the product of their
masses and inversely proportional to the square of the
distance between them.
OR mathematically,
2
1
rFg 21mmFg and
2
21
r
mmFg 2
21
r
mmGFg
force nalGravitatio:gF2 and 1 particle of masses:, 21 mm
2 and 1 particlebetween distance: r2211 kg m N x106.67 Constant nalgravitatio Universal: G
where
PHYSICS CHAPTER 7
6
The statement can also be shown by using the Figure 7.1.
where
2
211221
r
mmGFFF g
1m 2m
r12F
Figure 7.1
21F
2 particleon 1 particleby force nalGravitatio:12F
1 particleon 2 particleby force nalGravitatio:21F
Simulation 7.1
PHYSICS CHAPTER 7
7
Figures 7.2a and 7.2b show the gravitational force, Fg varies with the distance, r.
Notes:
Every spherical object with constant density can be reduced to a point mass at the centre of the sphere.
The gravitational forces always attractive in nature and the forces always act along the line joining the two point masses.
gF
r0
gF
2
1
r0
21mGmgradient
Figure 7.2a Figure 7.2b
PHYSICS CHAPTER 7
8
A spaceship of mass 9000 kg travels from the Earth to the Moon
along a line that passes through the Earths centre and the Moons
centre. The average distance separating Earth and the Moon is
384,000 km. Determine the distance of the spaceship from the
Earth at which the gravitational force due to the Earth twice the
magnitude of the gravitational force due to the Moon.
(Given the mass of the Earth, mE=6.001024 kg, the mass of the
Moon, mM=7.351022 kg and the universal gravitational constant,
G=6.671011 N m2 kg2)
Example 7.1 :
PHYSICS CHAPTER 7
9
Solution :
Given
kg; 107.35 kg; 106.00 22M24
E mm
m 103.84 kg; 0900 8EMs rm
Em Mmsm
x
EMr
xr EM
EsF
MsF
MsEs F2F
2EMsM
2
sE 2xr
mGm
x
mGm
ME
2
EM
2
2m
m
xr
x
2224
28
2
107.352
106.00
103.84
x
x
m 103.32 8x
PHYSICS CHAPTER 7
10
Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at
points A and B as shown in Figure 7.3. If a 50 g sphere is placed
at point C, determine
a. the resultant force acting on it.
b. the magnitude of the spheres acceleration.
(Given G = 6.671011 N m2 kg2)
Example 7.2 :
Figure 7.3
A B
C
cm 8kg 3.2 kg 2.5
g 50
cm 6
PHYSICS CHAPTER 7
11
Solution :
a.
The magnitude of the forces on mC,
22
311
2
AC
CAA
1010
10503.2106.67
r
mGmF
N 101.07 9AF
kg 1050kg; .52 kg; 3.2 3CBA mmm
m 1010m; 106 2AC2
BC rr
0.6sin 0.8cos
A B
C
m 108 2
m 10 6 2
m 10 10 2
AF
BF
PHYSICS CHAPTER 7
12
Solution :
22
311
2
BC
CBB
106
10502.5106.67
r
mGmF
N 10.322 9BF
Force xcomponent (N) ycomponent (N)
AF
F cosA F sinA
0.8101.07 910108.56
0.6101.07 910106.42
BF
BF09102.32
kg 1050kg; .52 kg; 3.2 3CBA mmm
m 1010m; 106 2AC2
BC rr
PHYSICS CHAPTER 7
13
Solution :
The magnitude of the nett force is
and its direction is
N 108.56 10 xF
22
yx FFF
N 10.96210.322106.42 9910 yF
29210 10.962108.56
N 10.083 9F
10
911
10.568
10.962tantan
x
y
F
F
.973 (254 from +x axis anticlockwise)
PHYSICS CHAPTER 7
14
Solution :
b. By using the Newtons second law of motion, thus
and the direction of the acceleration in the same direction of the
nett force on the mC i.e. 254 from +x axis anticlockwise.
amF C a39 1050103.08
28 s m 10.166 a
3
9
1050
103.08
a
PHYSICS CHAPTER 7
15
is defined as a region of space surrounding a body that has
the property of mass where the attractive force is experienced if a test mass placed in the region.
Field lines are used to show gravitational field around an object
with mass.
For spherical objects (such as the Earth) the field is radial as
shown in Figure 7.4.
7.1.2 Gravitational Field
M
Figure 7.4
PHYSICS CHAPTER 7
16
The gravitational field in small region near the Earths surfaceare uniform and can be drawn parallel to each other as shown in Figure 7.5.
The field lines indicate two things:
The arrows the direction of the field
The spacing the strength of the field
Figure 7.5
The gravitational field is a conservative field in which the work done in moving a body from one point to another is independent of the path taken.
Note:
New
PHYSICS CHAPTER 7
17
Exercise 7.1 :
Given G = 6.671011 N m2 kg2
1. Four identical masses of 800 kg each are placed at the corners
of a square whose side length is 10.0 cm. Determine the nett
gravitational force on one of the masses, due to the other three.
ANS. : 8.2103 N; 45
2. Three 5.0 kg spheres are located in the xy plane as shown in Figure 7.6.Calculate the magnitude
of the nett gravitational force
on the sphere at the origin due to
the other two spheres.
ANS. : 2.1108 N
Figure 7.6
PHYSICS CHAPTER 7
18
Exercise 7.1 :
3.
In Figure 8.7, four spheres form the corners of a square
whose side is 2.0 cm long. Calculate the magnitude and
direction of the nett gravitational force on a central sphere with
mass of m5 = 250 kg.
ANS. : 1.68102 N; 45
Figure 7.7
PHYSICS CHAPTER 7
19
At the end of this chapter, students should be able to:
Define gravitational field strength as gravitational force per unit mass,
Derive and use the equation for gravitational field strength.
Sketch a graph of ag against r and explain the change in agwith altitude and depth from the surface of the earth.
Learning Outcome:
7.1.3 Gravitational force and field strength
m
Fag
g
2r
MGa g
ww
w.k
ms.m
atr
ik.e
du
.my/p
hysic
s
PHYSICS CHAPTER 7
20
7.1.3 Gravitational field strength, is defined as the gravitational force per unit mass
of a body (test mass) placed at a point.
OR
It is a vector quantity.
The S.I. unit of the gravitational field strength is N kg1 or m s2.
ga
m
Fa
g
g
where
strength field nalGravitatio:ga
force nalGravitatio:gF
mass)(test body a of mass:m
PHYSICS CHAPTER 7
21
It is also known as gravitational acceleration (the freefall acceleration).
Its direction is in the same direction of the gravitational force.
Another formula for the gravitational field strength at a point is given by
m
Fa
g
g and 2g r
GMmF
2g r
GMa
masspoint and massst between te distance : r
2g
1
r
GMm
ma
where
masspoint theof mass :M
PHYSICS CHAPTER 7
22
Figure 7.8 shows the direction of the gravitational field strength
on a point S at distance r from the centre of the planet.
2r
GMag
r
M
Figure 7.8
PHYSICS CHAPTER 7
23
The gravitational field in the small region near the Earths
surface( r R) are uniform where its strength is 9.81 m s2 and its direction can be shown by using the Figure 7.9.
Figure 7.9
2R
GMgag
Earth theof radius :Rwhere2s m 9.81onaccelerati nalgravitatio : g
PHYSICS CHAPTER 7
24
Determine the Earths gravitational field strength
a. on the surface.
b. at an altitude of 350 km.
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.00 1024 kg and radius of the Earth, R = 6.40 106 m)
Solution :
a.
Example 7.3 :
RM
gaRr g m; 1040.66
26
2411
21040.6
1000.61067.6
R
GMg
The gravitational field strength is
1kg N 77.9 g OR 2s m 77.9
rg
(Towards the centre of the Earth)
PHYSICS CHAPTER 7
25
Solution :
b.
2g r
GMa
26
2411
1075.6
106.001067.6
2g s m 78.8
a(Towards the centre of the Earth)
R M
hRr 36 103501040.6
m 1075.6 6r
ga
h
r
The gravitational field strength is given by
PHYSICS CHAPTER 7
26
The gravitational field strength on the Earths surface is 9.81 N kg1.
Calculate
a. the gravitational field strength at a point C at distance 1.5R from
the Earths surface where R is the radius of the Earth.
b. the weight of a rock of mass 2.5 kg at point C.
Solution :
a. The gravitational field strength on the Earths surface is
The distance of point C from the Earths centre is
Example 7.4 :
1kg N 81.9 g
1
2kg N 81.9
R
GMg
RRRr 5.25.1
PHYSICS CHAPTER 7
27
Solution :
a. Thus the gravitational field strength at point C is given by
b. Given
The weight of the rock is
N 93.3W
2
Cr
GMag 25.2 R
GMag
225.6
1
R
GM
gmaW kg 5.2m
57.15.2
1kg N 57.181.925.6
1 ga
(Towards the centre of the Earth)
(Towards the centre of the Earth)
PHYSICS CHAPTER 7
28
Figure 8.10 shows an object A at a distance of 5 km from the object
B. The mass A is four times of the mass B. Determine the location
of a point on the line joining both objects from B at which the nett
gravitational field strength is zero.
Example 7.5 :
A
B
km 5
Figure 7.10
PHYSICS CHAPTER 7
29
Solution :
At point C,
BA3 4 m; 105 MMr
0nett
ga
2B
23
B
105
4
x
M
x
M
m 10.671 3x
r
ABC
xr x
2ga
1ga
21 gg aa
2B
2
A
x
GM
xr
GM
PHYSICS CHAPTER 7
30
Outside the Earth ( r > R)
Figure 8.11 shows a test mass which is outside the Earth and at
a distance r from the centre.
The gravitational field strength outside the Earth is
7.1.4 Variation of gravitational field strength on the
distance from the centre of the Earth
R
rM
Figure 8.11
2g r
GMa
2g
1
ra
PHYSICS CHAPTER 7
31
On the Earth ( r = R)
Figure 7.12 shows a test mass on the Earths surface.
The gravitational field strength on the Earths surface is
R
rM
Figure 7.12
2
2gs m 81.9 g
R
GMa
PHYSICS CHAPTER 7
32
R
r
M
'M
Inside the Earth ( r < R)
Figure 7.13 shows a test mass which is inside the Earth and at
distance r from the centre.
The gravitational field strength inside the Earth is given by
Figure 7.13
2g
'
r
GMa
where
portion spherical of mass the: 'M radius, ofEarth theof r
PHYSICS CHAPTER 7
33
By assuming the Earth is a solid sphere and constant
density, hence
Therefore the gravitational field strength inside the Earth is
V
V
M
M
''
3
3
3
34
3
34'
R
r
R
r
M
M
MR
rM
3
3
'
2
3
3
gr
MR
rG
a
rR
GMa
3g ra g
PHYSICS CHAPTER 7
34
The variation of gravitational field strength, ag as a function of
distance from the centre of the Earth, r is shown in Figure 7.14.
Figure 7.14
R
ga
r0 R
gR
GMa
2g
ra g2g
1
ra
PHYSICS CHAPTER 7
35
At the end of this chapter, students should be able to:
Define gravitational potential in a gravitational field.
Derive and use the formulae,
Sketch the variation of gravitational potential, V with distance, r from the centre of the earth.
Learning Outcome:
7.2 Gravitational potential ( hour)
r
GMV
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PHYSICS CHAPTER 7
36
7.2 Gravitational potential
7.2.1 Work done by the external force
Consider an external force, F
is required to bring a test
mass, m from r1 to r2 ,
as shown in Figure 7.18.
At the distance r2 from the
centre of the Earth,
The work done by the
external force through
the small displacement
dr is
m
M
1r2r
F
gF
dr
Figure 7.18
gFF
0cosFdrdW drFdW g
PHYSICS CHAPTER 7
37
Therefore the work done by the external force to bring test
mass, m from r1 to r2 is
2
1
r
rgdrFdW
2
12
r
rdr
r
GMmW
2r
GMmFg and
2
1
1r
rrGMmW
2
12
1r
rdr
rGMmW
21
1121 rr
GMmW rr
where
distance final: 2rdistance initial : 1r
PHYSICS CHAPTER 7
38
at a point is defined as the work done by an external force in bringing a test mass from infinity to a point per unit the test mass.
OR mathematically, V is written as:
It is a scalar quantity.
Its dimension is given by
7.2.2 Gravitational potential, V
m
WV
where
mass test theof mass :m point aat potential nalgravitatio :V
mass test a bringingin done work :W point a oinfinity t from
mW
V M
TML 22 V
22TL V
PHYSICS CHAPTER 7
39
The S.I unit for gravitational potential is m2 s2 or J kg1.
Another formula for the gravitational potential at a point is given by
21
11
rrm
GMmV
m
WV and
21
11
rrGMmW
where 1r and rr 2
rm
GMmV
11
r
GMV
where
point ebetween th distance : rM mass,point theand
PHYSICS CHAPTER 7
40
The gravitational potential difference between point A and B
(VAB) in the Earths gravitational field is defined as the work done in bringing a test mass from point B to point A per
unit the test mass.
OR mathematically, VAB is written as:
BABA
AB VVm
WV
where
A.point toBpoint from
mass test thebringingin done work :BAW
Apoint at potential nalgravitatio : AV
Bpoint at potential nalgravitatio : BV
PHYSICS CHAPTER 7
41
Figure 7.19 shows two points A and B at a distance rA and rBfrom the centre of the Earth respectively in the Earths gravitational field.
M
A
BrA
rB
Figure 7.19
The gravitational potential
difference between the points A
and B is given by
BAAB VVV
BA
ABr
GM
r
GMV
AB
AB
11
rrGMV
PHYSICS CHAPTER 7
42
The gravitational potential difference between point B and A in
the Earths gravitational field is given by
The variation of gravitational potential, V when the test mass, mmove away from the Earths surface is illustrated by the graph in Figure 7.20.
m
WVVV ABABBA
R
R
GM
r0
V
rV
1
Note:
The Gravitational potential at infinity is zero. 0V
Figure 7.20
PHYSICS CHAPTER 7
43
When in orbit, a satellite attracts the Earth with a force of 19 kN
and the satellites gravitational potential due to the Earth is 5.45107 J kg1.
a. Calculate the satellites distance from the Earths surface.
b. Determine the satellites mass.
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 5.981024 kg and radius of the Earth , R = 6.38106 m)
Solution :
Example 7.7 :
R
gF
rh
173 kg J 10455 N; 1019 .VFg
PHYSICS CHAPTER 7
44
Solution :
a. By using the formulae of gravitational potential, thus
Therefore the satellites distance from the Earths surface is
r
GMV
m 1032.7 6r
66 1038.61032.7 h
m 104.9 5h
r
.. 24117 10985106761045.5
Rhr
173 kg J 10455 N; 1019 .VFg
PHYSICS CHAPTER 7
45
Solution :
b. From the Newtons law of gravitation, hence
2r
GMmFg
kg 2552m
26
24113
10327
10985106761019
.
m..
173 kg J 10455 N; 1019 .VFg
PHYSICS CHAPTER 7
46
At the end of this chapter, students should be able to:
Explain satellite motion with:
velocity,
period,
Learning Outcome:
7.3 Satellite motion in a circular orbit ( hour)
r
GMv
GM
rT
3
2
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47
7.3 Satellite motion in a circular orbit7.3.1 Tangential (linear/orbital) velocity, v
Consider a satellite of mass, m travelling around the Earth of mass, M, radius, R, in a circular orbit of radius, r with constant tangential (orbital) speed, v as shown in Figure 7.22.
Figure 7.22
PHYSICS CHAPTER 7
48
The centripetal force, Fc is contributed by the gravitational force
of attraction, Fg exerted on the satellite by the Earth.
Hence the tangential velocity, v is given by
ccg maFF
r
mv
r
GMm 2
2
r
GMv
where
Earth theof mass :M
from satellite theof distance :rEarth theof centre the
constant nalgravitatio universal : G
PHYSICS CHAPTER 7
49
For a satellite close to the Earths surface,
Therefore
The relationship between tangential velocity and angular
velocity is
Hence , the period, T of the satellite orbits around the Earth is given by
Rr and 2gRGM
gRv
T
rrv
2
r
GM
T
r2
GM
rT
3
2
PHYSICS CHAPTER 7
50
Figure 8.23 shows a synchronous (geostationary) satellite which stays above the same point on the equator of the Earth.
The satellite have the following characteristics:
It revolves in the same direction as the Earth.
It rotates with the same period of rotation as that of the Earth (24 hours).
It moves directly above the equator.
The centre of a synchronous satellite orbit is at the centre of the Earth.
It is used as a communication satellite.
7.3.2 Synchronous (Geostationary) Satellite
Figure 8.23
PHYSICS CHAPTER 7
51
The weight of a satellite in a circular orbit round the Earth is half of
its weight on the surface of the Earth. If the mass of the satellite is
800 kg, determine
a. the altitude of the satellite,
b. the speed of the satellite in the orbit,
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.001024 kg, and radius of the Earth , R = 6.40106 m)
Example 7.12 :
PHYSICS CHAPTER 7
52
Solution :
a. The satellite orbits the Earth in the circular path, thus
b. The speed of the satellite is given by
PHYSICS CHAPTER 7
53
The radius of the Moons orbit around the Earth is 3.8 108 m and the period of the orbit is 27.3 days. The masses of the Earth and
Moon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate
the total energy of the Moon in the orbit.
Solution :
The period of the satellite is
The tangential speed of the satellite is
Example 7.13 :
s m 50.9 kg; 120 m; 1050.8 26 gmr
T
rv
2
13 s m 1024.4 v
36005.3Ts 12600T
12600
1050.82 6
v
PHYSICS CHAPTER 7
54
Solution :
A satellite orbits the planet in the circular path, thus
cg FF
2
2 r
mv
r
GMm
s m 50.9 kg; 120 m; 1050.8 26 gmr
r
GMv 2 and
2gRGM
r
gRv
22
6
223
1050.8
50.91024.4
R
m 1001.4 6R
PHYSICS CHAPTER 7
55
Exercise 7.2 :
Given G = 6.671011 N m2 kg2
1. A rocket is launched vertically from the surface of the Earth
at speed 25 km s1. Determine its speed when it escapes from
the gravitational field of the Earth.
(Given g on the Earth = 9.81 m s2, radius of the Earth ,
R = 6.38 106 m)
ANS. : 2.24104 m s1
2. A satellite revolves round the Earth in a circular orbit whose
radius is five times that of the radius of the Earth. The
gravitational field strength at the surface of the Earth is
9.81 N kg1. Determine
a. the tangential speed of the satellite in the orbit,
b. the angular frequency of the satellite.
(Given radius of the Earth , R = 6.38 106 m)
ANS. : 3538 m s1 ; 1.11104 rad s1
PHYSICS CHAPTER 7
56
Exercise 7.2 :
3. A geostationary satellite of mass 2400 kg is placed
35.92 Mm from the Earths surface orbits the Earth along a circular path.
Determine
a. the angular velocity of the satellite,
b. the tangential speed of the satellite,
c. the acceleration of the satellite,
d. the force of attraction between the Earth and the satellite,
e. the mass of the Earth.
(Given radius of the Earth , R = 6.38 106 m)
ANS. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2;
537 N ; 6.001024 kg
PHYSICS CHAPTER 7
57
THE ENDNext Chapter
CHAPTER 8 :
Simple Harmonic Motion
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1PHYSICS CHAPTER 8
CHAPTER 8: CHAPTER 8: Rotational of rigid bodyRotational of rigid body
(8 Hours)(8 Hours)
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PHYSICS CHAPTER 8
2
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: a) Define and describea) Define and describe::
angular displacement (angular displacement ()) average angular velocity (average angular velocity (avav)) instantaneous angular velocity (instantaneous angular velocity ()) average angular acceleration (average angular acceleration (avav) ) instantaneous angular acceleration (instantaneous angular acceleration (). ).
b) Relate b) Relate parameters in rotational motion with their corresponding parameters in rotational motion with their corresponding quantities in linear motion.quantities in linear motion. Write and use Write and use ::
c) c) UseUse equations for rotational motion with constant angular equations for rotational motion with constant angular acceleration.acceleration.
Learning Outcome:
8.1 Rotational Kinematics (2 hour)
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s=r ; v=r ;a t=r ;ac=r2= v
2
r
=0t =0 t12 t
2 2=022
PHYSICS CHAPTER 8
3
8.1 Parameters in rotational motion8.1.1 Angular displacement, is defined as an angle through which a point or line has an angle through which a point or line has
been rotated in a specified direction about a specified axis.been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad)radian (rad). Figure 8.1 shows a point P on a rotating compact disc (CD)
moves through an arc length s on a circular path of radius r about a fixed axis through point O.
Figure 8.1Figure 8.1
PHYSICS CHAPTER 8
4
From Figure 8.1, thus
Others unit for angular displacement is degree (degree ()) and revolution (rev)revolution (rev). Conversion factor :
Sign convention of angular displacement : PositivePositive if the rotational motion is anticlockwiseanticlockwise. NegativeNegative if the rotational motion is clockwiseclockwise.
1 rev=2 rad=360
= sr OR s=rwhere : angle angular displacement in radian
s : arc lengthr : radius of the circle
PHYSICS CHAPTER 8
5
Average angular velocity, Average angular velocity, avav is defined as the rate of change of angular displacementthe rate of change of angular displacement. Equation :
Instantaneous angular velocity, Instantaneous angular velocity, is defined as the instantaneous rate of change of angular the instantaneous rate of change of angular
displacementdisplacement. Equation :
8.1.2 Angular velocity
av=21t2t1
= t
= limitt 0
t =
ddt
where 2 : final angular displacement in radian
t : time interval1 : initial angular displacement in radian
PHYSICS CHAPTER 8
6
It is a vector quantityvector quantity. The unit of angular velocity is radian per second (rad sradian per second (rad s11)) Others unit is revolution per minute (rev minrevolution per minute (rev min11 or rpm) or rpm)
Conversion factor:
Note : Every partEvery part of a rotating rigid body has the same angular same angular
velocityvelocity.Direction of the angular velocityDirection of the angular velocity Its direction can be determine by using right hand grip ruleright hand grip rule
where
1 rpm =260 rad s1=
30 rad s
1
ThumbThumb : direction of angular velocityangular velocityCurl fingersCurl fingers : direction of rotationrotation
PHYSICS CHAPTER 8
7
Figures 8.2 and 8.3 show the right hand grip rule for determining the direction of the angular velocity.
Figure 8.2Figure 8.2
Figure 8.3Figure 8.3
PHYSICS CHAPTER 8
8
The angular displacement, of the wheel is given by
where in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine
a. the angle, in degree, at time 2.2 s and 4.8 s,b. the distance that a particle on the rim moves during that time interval,c. the average angular velocity, in rad s1 and in rev min1 (rpm), between 2.2 s and 4.8 s,d. the instantaneous angular velocity at time 3.0 s.
Example 8.1 :
=5t2 t
PHYSICS CHAPTER 8
9
Solution :Solution :
a. At time, t1 =2.2 s :
At time, t2 =4.8 s :
r=d2=0 .56
2 =0 . 28 m
1=5 2. 22 2.2
1=22 rad
2=5 4 .824 .8
2=110 rad
PHYSICS CHAPTER 8
10
Solution :Solution :
b. By applying the equation of arc length,
Therefore
c. The average angular velocity in rad s1 is given by
r=d2=0 .56
2 =0 . 28 m
s=rs=r=r 21 s=0 . 28 11022
av=t=21t2t1
av=11022 4 . 82. 2
PHYSICS CHAPTER 8
11
Solution :Solution :c. and the average angular velocity in rev min1 is
d. The instantaneous angular velocity as a function of time is
At time, t =3.0 s :
av=33 . 9 rad1 s 1 rev2 rad 60 s1 min
= ddt5t2t
=ddt
=10 t1
=10 3 .0 1
PHYSICS CHAPTER 8
12
A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the divers average angular (rotational) velocity during a dive.(Given g = 9.81 m s2)Solution :Solution :
Example 8.2 :
u y=00=0
10 m
water1=2. 5 rev
PHYSICS CHAPTER 8
13
Solution :Solution :From the diagram,Thus
Therefore the divers average angular velocity is
1=2.52=5 rads y=10 m
s y=u y t12 gt
2
10=0129 . 81 t 2
av=10
t
av=501 . 43
PHYSICS CHAPTER 8
14
Average angular acceleration, Average angular acceleration, avav is defined as the rate of change of angular velocitythe rate of change of angular velocity. Equation :
Instantaneous angular acceleration, Instantaneous angular acceleration, is defined as the instantaneous rate of change of angular the instantaneous rate of change of angular
velocityvelocity. Equation :
8.1.3 Angular acceleration
av=21t2t1
= t
= limitt 0
t =
ddt
where 2 : final angular velocity
t : time interval1 : initial angular velocity
PHYSICS CHAPTER 8
15Figure 8.4Figure 8.4
It is a vector quantityvector quantity. The unit of angular acceleration is rad srad s22. Note:
If the angular acceleration, is positivepositive, then the angular velocity, is increasingincreasing.
If the angular acceleration, is negativenegative, then the angular velocity, is decreasingdecreasing.
Direction of the angular accelerationDirection of the angular acceleration If the rotation is speeding upspeeding up, and in the same directionsame direction
as shown in Figure 8.4.
PHYSICS CHAPTER 8
16
Figure 8.5Figure 8.5
If the rotation is slowing downslowing down, and have the opposite opposite directiondirection as shown in Figure 8.5.
Example 8.3 :The instantaneous angular velocity, of the flywheel is given by
where in radian per second and t in seconds. Determinea. the average angular acceleration between 2.2 s and 4.8 s,b. the instantaneous angular acceleration at time, 3.0 s.
=8t3t 2
PHYSICS CHAPTER 8
17
Solution :Solution :a. At time, t1 =2.2 s :
At time, t2 =4.8 s :
Therefore the average angular acceleration is
1=8 2 . 2 32 . 22
1=80 . 3 rad s1
2=8 4 .8 34 . 8 2
av=21t 2t1
av=86280 .34 . 82. 2
PHYSICS CHAPTER 8
18
Solution :Solution :b. The instantaneous angular acceleration as a function of time is
At time, t =3.0 s :
= ddt8t3 t2
= ddt
=24 3 . 0 22 3 . 0
PHYSICS CHAPTER 8
19
Exercise 8.1 :1. If a disc 30 cm in diameter rolls 65 m along a straight line
without slipping, calculatea. the number of revolutions would it makes in the process,b. the angular displacement would be through by a speck of
gum on its rim.ANS. : 69 rev; 138ANS. : 69 rev; 138pipi rad rad2. During a certain period of time, the angular displacement of a
swinging door is described by
where is in radians and t is in seconds. Determine the angular displacement, angular speed and angular accelerationa. at time, t =0,b. at time, t =3.00 s.
ANS. : ANS. : 5.00 rad, 10.0 rad s5.00 rad, 10.0 rad s11, 4.00 rad s, 4.00 rad s22; 53.0 rad, 22.0 rad s; 53.0 rad, 22.0 rad s11, , 4.00 rad s4.00 rad s22
=5 . 0010 . 0t2. 00 t2
PHYSICS CHAPTER 8
20
8.1.2 Relationship between linear and rotational motion 8.1.2 Relationship between linear velocity, v and
angular velocity, When a rigid body is rotates about rotation axis O , every
particle in the body moves in a circle as shown in the Figure 8.6.
v
s
y
x
rP
O
Figure 8.6Figure 8.6
PHYSICS CHAPTER 8
21
Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by
The directiondirection of the linear (tangential) velocitylinear (tangential) velocity always tangent to the circular pathtangent to the circular path.
Every particle on the rigid body has the same angular speedsame angular speed (magnitude of angular velocity) but the tangential speedtangential speed is notnot the samesame because the radiusradius of the circle, rr is changing changing dependdepend on the position of the particleposition of the particle.
v=dsdtv=r ddt
s=r
v=r
and
Simulation 7.1
PHYSICS CHAPTER 8
22
at
aca
x
y
P
O
If the rigid bodyrigid body is gaining the angular speedgaining the angular speed then the tangential velocitytangential velocity of a particle also increasingincreasing thus twotwo component of accelerationacceleration are occurredoccurred as shown in Figure 8.7.
8.1.2 Relationship between tangential acceleration, at and angular acceleration,
Figure 8.7Figure 8.7
PHYSICS CHAPTER 8
23
The components are tangential acceleration, tangential acceleration, aatt and centripetal acceleration, centripetal acceleration, aacc given by
but
The vector sum of centripetal and tangential accelerationvector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, resultant (linear) acceleration, aa given by
and its magnitude,
a t=dvdt
a t=rddt
a t=r
v=rand
ac=v2
r=r2=v
a=a taca=a t2ac2
Vector formVector form
PHYSICS CHAPTER 8
24
8.1.3 Rotational motion with uniform angular acceleration Table 8.1 shows the symbols used in linear and rotational
kinematics.
Table 8.1Table 8.1
Linear motion Quantity
Rotational motion
s DisplacementDisplacementu 0Initial velocityInitial velocityv Final velocityFinal velocitya AccelerationAccelerationt tTimeTime
PHYSICS CHAPTER 8
25
Table 8.2 shows the comparison of linear and rotational motion with constant acceleration.
Linear motion Rotational motion
a=constant
v=uat
=constant=0t
s=ut12 at2 =0 t
12 t
2
v2=u22 as 2=022
s=12vu t =12 0 t
where in radian. Table 8.2Table 8.2
PHYSICS CHAPTER 8
26
A car is travelling with a velocity of 17.0 m s1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s2 for 5.00 s, calculatea. the number of revolutions of the wheels during this period,b. the angular speed of the wheels after 5.00 s.Solution :Solution :a. The initial angular velocity is
and the angular acceleration of the wheels is given by
Example 8.4 :
u=17 . 0 m s1 , r=0 . 48 m , a=2 . 00 m s2 , t=5 . 00 su=r0
17 . 0=0 . 48 0
2 . 00=0 . 48 a=r
PHYSICS CHAPTER 8
27
Solution :Solution :a. By applying the equation of rotational motion with constant angular acceleration, thus
therefore
b. The angular speed of the wheels after 5.00 s is
=0 t12 t
2
=229 rad
u=17 . 0 m s1 , r=0 . 48 m , a=2 . 00 m s2 , t=5 . 00 s
=35 . 4 5 . 0012 4 . 17 5 . 002
=0t=35 . 44 . 17 5 . 00
PHYSICS CHAPTER 8
28
The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h1 to 35.0 km h1. The wheels have a diameter of 70 cm.a. Calculate the angular acceleration.b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. Solution :Solution :
Example 8.5 :
=302=60 rad , r=0 .702 =0 . 35 m ,
u=50 . 0 km1 h 10
3 m1 km 1 h3600 s =13 . 9 m s1 ,
v=35 .0 km1 h 10
3 m1 km 1 h3600 s =9 . 72 m s1
PHYSICS CHAPTER 8
29
Solution :Solution :a. The initial angular speed of the wheels is
and the final angular speed of the wheels is
therefore
b. The car stops thus Hence
u=r013 .9=0 .35 0
v=r9 . 72=0 . 35
2=022
27 . 8 2=39 . 722 60
=0 0=27 .8 rad s1
and=0t0=27 .8 2. 13 t
PHYSICS CHAPTER 8
30
A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s1. The angular acceleration of the blade is 0.750 rev s2. Determinea. the angular velocity after 4.00 s,b. the number of revolutions for the blade turns in this time interval,c. the tangential speed of a point on the tip of the blade at time, t =4.00 s,d. the magnitude of the resultant acceleration of a point on the tip of the blade at t =4.00 s.Solution :Solution :
a. Given t =4.00 s, thus
Example 8.6 :
r=0 . 400 m , 0=0. 1502=0 .300 rad s1 ,
=0 .7502=1. 50 rad s2
=0t=19 . 8 rad s1= 0 . 3001 .50 4 .00
PHYSICS CHAPTER 8
31
Solution :Solution :b. The number of revolutions of the blade is
c. The tangential speed of a point is given by
=0 t12 t
2
=41. 5 rad=0 . 300 4 . 0012 1 .50 4 . 00
2
v=rv=0 . 400 19 . 8
PHYSICS CHAPTER 8
32
Solution :Solution :d. The magnitude of the resultant acceleration is
a=ac2at2
a=v 2r 2r 2a=7 .9220 . 400 2 0 . 4001. 50 2
PHYSICS CHAPTER 8
33
Calculate the angular velocity ofa. the secondhand,b. the minutehand andc. the hourhand,of a clock. State in rad s1.d. What is the angular acceleration in each case?Solution :Solution :a. The period of secondhand of the clock is T = 60 s, hence
Example 8.7 :
=2T =260
PHYSICS CHAPTER 8
34
Solution :Solution :b. The period of minutehand of the clock is T = 60 min = 3600 s, hence
c. The period of hourhand of the clock is T = 12 h = 4.32 104 s, hence
d. The angular acceleration in each cases is
=23600
=24 . 32104
PHYSICS CHAPTER 8
35
A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s2, calculate the distance travelled by the coin before coming to rest.Solution :Solution :
The radius of the coin is
Example 8.8 :
d=2 . 40102 m
0=18 rad s1
s
=1 .90 rad s2=0 rad s1
r=d2=1 .20102 m
PHYSICS CHAPTER 8
36
Solution :Solution :The initial speed of the point at the edge the coin is
and the final speed isThe linear acceleration of the point at the edge the coin is given by
Therefore the distance travelled by the coin is
u=r0u=1 . 20102 18
v=0 m s1
a=ra=1. 20102 1. 90
v2=u22as0=0 .216 22 2 .28102 s
PHYSICS CHAPTER 8
37
Exercise 8.2 :1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev
min1 about its central axis. Determinea. its angular speed,b. the tangential speed at a point 3.00 cm from its centre,c. the radial acceleration of a point on the rim,d. the total distance a point on the rim moves in 2.00 s.
ANS. : ANS. : 126 rad s126 rad s11; 3.77 m s; 3.77 m s11; 1.26 ; 1.26 10 1033 m s m s22; 20.1 m; 20.1 m
2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculatea. its angular velocity in rad s1,b. the linear speed and the radial acceleration of a point on the
edge of the grinding wheel.ANS. : ANS. : 262 rad s262 rad s11; 46 m s; 46 m s11, 1.2 , 1.2 10 1044 m s m s22
PHYSICS CHAPTER 8
38
Exercise 8.2 :3. A rotating wheel required 3.00 s to rotate through 37.0
revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s1. Calculate the constant angular acceleration of the wheel.
ANS. : ANS. : 13.6 rad s13.6 rad s22
4. A wheel rotates with a constant angular acceleration of 3.50 rad s2.a. If the angular speed of the wheel is 2.00 rad s1 at t =0, through what angular displacement does the wheel rotate in 2.00 s.b. Through how many revolutions has the wheel turned during this time interval?c. What is the angular speed of the wheel at t = 2.00 s?
ANS. : ANS. : 11.0 rad; 1.75 rev; 9.00 rad s11.0 rad; 1.75 rev; 9.00 rad s11
PHYSICS CHAPTER 8
39
Exercise 8.2 :5. A bicycle wheel is being tested at a repair shop. The angular
velocity of the wheel is 4.00 rad s1 at time t = 0 , and its angular acceleration is constant and equal 1.20 rad s2. A spoke OP on the wheel coincides with the +xaxis at time t = 0 as shown in Figure 8.8.
a. What is the wheels angular velocity at t = 3.00 s?b. What angle in degree does the spoke OP make with the positive xaxis at this time?
ANS. :ANS. : 0.40 rad s0.40 rad s11; 18; 18
Figure 8.8Figure 8.8
x
y
PO
PHYSICS CHAPTER 8
40
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and use Define and use torque. torque.
State and use State and use conditions for equilibrium of rigid body:conditions for equilibrium of rigid body:
Learning Outcome:
8.2 Equilibrium of a uniform rigid body (2 hour)
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=== 0 , 0 , 0 FF yx
PHYSICS CHAPTER 8
41
8.2.1 Torque (moment of a force), The magnitude of the torquemagnitude of the torque is defined as the product of a the product of a
force and its perpendicular distance from the line of action force and its perpendicular distance from the line of action of the force to the point (rotation axis)of the force to the point (rotation axis).OR
Because ofwhere r : distance between the pivot point (rotation
axis) and the point of application of force.Thus
Fd =
force theof magnitude :Farm)(moment distancelar perpendicu : d
torque theof magnitude : where
sinrd =
sin Fr=rF
and between angle : where
OR Fr
=
PHYSICS CHAPTER 8
42
It is a vector quantityvector quantity. The dimension of torque is
The unit of torqueunit of torque is N mN m (newton metre), a vector productvector product unlike the joule (unit of work)joule (unit of work), also equal to a newton metre, which is scalar productscalar product.
Torque is occurred because of turning (twisting) effects of turning (twisting) effects of the forces the forces on a body.
Sign convention of torque: PositivePositive  turning tendency of the force is anticlockwiseanticlockwise. NegativeNegative  turning tendency of the force is clockwiseclockwise.
The value of torque dependstorque depends on the rotation axisrotation axis and the magnitude of applied forcemagnitude of applied force.
[ ] [ ][ ] 22TMLdF ==
PHYSICS CHAPTER 8
43
Case 1 :Case 1 : Consider a force is applied to a metre rule which is pivoted at
one end as shown in Figures 8.9a and 8.9b.
Figure 8.9aFigure 8.9a
F
F
Figure 8.9bFigure 8.9b
Pivot point (rotation axis)
Fd =
rd sin=
FrFd sin==
(anticlockwise)
(anticlockwise)r
Point of action of a force
Line of action of a force
d
PHYSICS CHAPTER 8
44
O
Figure 8.10Figure 8.102
Case 2 :Case 2 : Consider three forces are applied to the metre rule which is
pivoted at one end (point O) as shown in Figures 8.10.
Caution : If the line of action of a force is through the rotation axisline of action of a force is through the rotation axis
then
1F
1
111 rd sin=
321 ++= OTherefore the resultant (nett) torque is
3F
2F 1r
0sin === 333333 rFdF
222 rd sin=
111111 rFdF sin==222222 rFdF sin==
2r
2211 dFdF = OFr sin=
0=and 0=
Simulation 5.1
PHYSICS CHAPTER 8
45
Determine a resultant torque of all the forces about rotation axis, O in the following problems.a.
Example 8.9 :
m 5
N 10=2Fm 5 N 30=1F
m 3
m 3
N 20=3F
m 10
m 6O
PHYSICS CHAPTER 8
46
b.Example 8.9 :
m 5
N 10=2F
m 5
N 30=1F
m 3
m 3
N 25=4F
N 20=3F
m 10
m 6O
PHYSICS CHAPTER 8
47
m 5m 5
m 10
m 6O
Solution :Solution :a.
Force Torque (N m), o=Fd=Frsin
1F ( ) ( ) 90330 =
2F ( )( ) 50510 +=+
N 10=2FN 30=1F
N 20=3F
m 3=1d
m 5=2d
3F
0The resultant torque:
PHYSICS CHAPTER 8
48
m 5
m 10
m 3
m 6
m 5
Solution :Solution :b.
Force Torque (N m), o=Fd=Frsin
1F ( ) ( ) 90330 =
2F
( )( )( ) 51.50.515520sin ==rF33F
0 The resultant torque:
N 10=2F
N 30=1F
0.51553
3sin22
=
+=
O
N 20=3F
N 25=4F
m 3=1dm 5=r
4F
0
3d
PHYSICS CHAPTER 8
49
8.2 Equilibrium of a rigid body8.2.1.1 Nonconcurrent forces is defined as the forces whose lines of action do not pass the forces whose lines of action do not pass
through a single common point. through a single common point. The forces cause the rotational motionrotational motion on the body. The combination of concurrent and nonconcurrent forces cause
rolling motionrolling motion on the body. (translational and rotationaltranslational and rotational motion)
Figure 8.11 shows an example of nonconcurrent forces.
2F
3F
1F
Figure 8.11Figure 8.11
4F
PHYSICS CHAPTER 8
50
8.2.1.2 Equilibrium of a rigid body Rigid bodyRigid body is defined as a body with definite shape that a body with definite shape that
doesnt change, so that the particles that compose it stay in doesnt change, so that the particles that compose it stay in fixed position relative to one another even though a force is fixed position relative to one another even though a force is exerted on itexerted on it.
If the rigid body is in equilibriumrigid body is in equilibrium, means the body is translational and rotational equilibriumtranslational and rotational equilibrium.
There are two conditionstwo conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must The vector sum of all forces acting on a rigid body must
be zero.be zero. == 0nettFFOR
=== 0 , 0 , 0 zyx FFF
PHYSICS CHAPTER 8
51
The vector sum of all external torques acting on a rigid The vector sum of all external torques acting on a rigid body must be zero about any rotation axisbody must be zero about any rotation axis.
This ensures rotational equilibriumrotational equilibrium. This is equivalent to the three independent scalar
equations along the direction of the coordinate axes,
Centre of gravity, CG Centre of gravity, CG is defined as the point at which the whole weight of a body the point at which the whole weight of a body
may be considered to actmay be considered to act. A force that exerts on the centre of gravityexerts on the centre of gravity of an object will
cause a translational motiontranslational motion.
== 0nett
=== 0 , 0 , 0 zyx
PHYSICS CHAPTER 8
52
Figures 8.14 and 8.15 show the centre of gravity for uniformcentre of gravity for uniform (symmetric) objectobject i.e. rod and sphere
rodrod refer to the midway point between its endmidway point between its end.
spheresphere refer to geometric centregeometric centre.
2l
2l
CG
CGl
Figure 8.12Figure 8.12
Figure 8.13Figure 8.13
PHYSICS CHAPTER 8
53
8.2.4 Problem solving strategies for equilibrium of a rigid body
The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagramSketch a simple diagram of the system to help
conceptualize the problem. Sketch a separate free body diagramSketch a separate free body diagram for each body. Choose a convenient coordinate axesChoose a convenient coordinate axes for each body and
construct a tableconstruct a table to resolve the forces into their components and to determine the torque by each force.
Apply the condition for equilibrium of a rigid bodyApply the condition for equilibrium of a rigid body :
SolveSolve the equationsequations for the unknowns.
= 0xF = 0yF; and = 0
PHYSICS CHAPTER 8
54
A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in Figure 8.14. If the mass of the beam is 3.0 kg, calculatea. the weight, W1 needed,b. the force exerted on the beam at point O.(Given g =9.81 m s2)
Example 8.10 :
1W2W
A BO35 cm 75 cm
Figure 8.14Figure 8.14
PHYSICS CHAPTER 8
55
Solution :Solution :The free body diagram of the beam :
Let point O as the rotation axis.
N 23 ;kg 3 == 2Wm
0.75 mA B
OCG
1W
2W
N
gm
0.35 m
0.55 m 0.55 m
0.20 m0.20 m
Force ycomp. (N) Torque (N m), o=Fd=Frsin1W
1W
gm ( )( )9.813 ( ) ( ) 5.880.2029.4 =
( ) 11 WW 0.750.75 =2W
23 ( )( ) 8.050.3523 =+
N
N 029.4=
PHYSICS CHAPTER 8
56
Solution :Solution :Since the beam remains at rest thus the system in equilibrium. a. Hence
b. 0= yFand
= 0O05.888.050.75 =+ 1W
029.423 =+ NW1( ) 029.4232.89 =+ N
PHYSICS CHAPTER 8
57
A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 8.15. The height of the end A of the ladder is 8.0 m from the rough floor.a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B.b. If the ladder is just on the verge of slipping when the firefighter is 7.0 m up the ladder , Calculate the coefficient of static friction between ladder and floor.(Given g =9.81 m s2)
Example 8.11 :
A
B
smooth wall
rough floorFigure 8.15Figure 8.15
PHYSICS CHAPTER 8
58
Solution :Solution :a. The free body diagram of the ladder : Let point B as the rotation axis.
kg 60 ;kg 5.0 == fl mm
A
B
CG
gm f
1N
gml
2N
m 8.0m 10
m 3.0
m 5.0
Force xcomp. (N)ycomp.
(N)Torque (N m),
B=Fd=Frsingml
1N1N
0.8108sin ==
sf
gm f
49.1 0.6106sin ==
2N
sf
0
5890
2N
0
0
m 6.0
( ) ( ) sin5.049.1
147=
0
( ) ( ) sin3.05891060= ( ) N1 sin10
1N8=0
0 sf
PHYSICS CHAPTER 8
59
Solution :Solution : Since the ladder in equilibrium thus
0= B081060147 =+ 1N
N 151=1N0= xF0= s1 fN
Horizontal force:Horizontal force:
0= yF058949.1 =+ 2N
Vertical force:Vertical force:
PHYSICS CHAPTER 8
60
m 10
A
B
m 8.0
m 6.0
m 5.0
Solution :Solution :b. The free body diagram of the ladder : Let point B as the rotation axis.
0.6sin 0.8;sin ==
gm f
gml
sf
m 7.0
Force xcomp. (N)ycomp.
(N)Torque (N m),
B=Fd=Frsingml
1N1N
2s N
gm f
49.1
2N
sf
0
5890
2N
0
0
( ) ( ) sin5.049.1
147=
0
( ) ( ) sin7.05892474= ( ) N1 sin10
1N8=0
0
2N
1N
PHYSICS CHAPTER 8
61
Solution :Solution : Consider the ladder stills in equilibrium thus
0= B082474147 =+ 1N
N 328=1N
0= xF0= 2s1 NN
0= yF058949.1 =+ 2N
N 638=2N
( ) ( ) 0638328 = s
PHYSICS CHAPTER 8
62
Figure 8.16Figure 8.16
A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 8.16. A cable at an angle 30 with the beam helps to support the light.a. Sketch a free body diagram of the beam.b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole.(Given g =9.81 m s2)
Example 8.12 :
PHYSICS CHAPTER 8
63
Solution :Solution :a. The free body diagram of the beam :
b. Let point O as the rotation axis.
kg 10.0 ;kg 20.0 == bf mm
Force xcomp. (N) ycomp. (N) Torque (N m), o=Fd=Frsingm f ( )l1960 196
O CG
gm f
gmb
T
S
30
ll0.5
gmb ( )( ) ll 49.10.598.1 =0 98.1
T
TlTl 0.530sin =30cosT 30sinT
S
xS yS 0
PHYSICS CHAPTER 8
64
Solution :Solution :b. The floodlight and beam remain at rest thus i.
ii.
0= O00.549.1196 =+ Tlll
0= xF0cos =+ xS30T
N 424=xS0= yF030sin98.1196 =++ yST
N 49.1=yS
PHYSICS CHAPTER 8
65
Solution :Solution :b. ii. Therefore the magnitude of the force is
and its direction is given by
2y
2x SSS +=
( ) ( ) 22S 49.1424 +=
=
x
y
SS
1tan
=
42449.1tan 1
PHYSICS CHAPTER 8
66
Exercise 8.3 :Use gravitational acceleration, g = 9.81 m s21.
Figure 8.17 shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle is 30. Calculate the resultant torque about point D.
ANS. : 3.7 N mANS. : 3.7 N m
D
A B
C
1F
3F
2F
Figure 8.17Figure 8.17
a
b
PHYSICS CHAPTER 8
67
Figure 8.18Figure 8.18
Exercise 8.3 :2.
A seesaw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 8.18. The fulcrum is under the centre of gravity of the board. Determinea. the magnitude of the force exerted by the fulcrum on the
board,b. where the father should sit from the fulcrum to balance the system.
ANS. : 1128 N; 1.31 mANS. : 1128 N; 1.31 m
PHYSICS CHAPTER 8
68
3.
A traffic light hangs from a structure as show in Figure 8.19. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole.
ANS. : 248 N; 197 N, 248 NANS. : 248 N; 197 N, 248 N
Figure 8.19Figure 8.19
Exercise 8.3 :
PHYSICS CHAPTER 8
69
4.
A uniform 10.0 N picture frame is supported by two light string as shown in Figure 8.20. The horizontal force, F is applied for holding the frame in the position shown.a. Sketch the free body diagram of the picture frame.b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F .
ANS. : 1.42 N, 11.2 N; 7.20 NANS. : 1.42 N, 11.2 N; 7.20 N
Exercise 8.3 :
Figure 8.20Figure 8.20
F
50.0
cm 15.0
cm 30.0
PHYSICS CHAPTER 8
70
Learning Outcome:
8.3 Rotational dynamics (1 hour)
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define Define the moment of inertia of a rigid body about an axis,the moment of inertia of a rigid body about an axis,
State and useState and use torque, torque,
=
=
n
1iiirmI2
I =
PHYSICS CHAPTER 8
71
8.3.1 Centre of mass, moment of inertia and torque8.3.1.1 Centre of mass (CM) is defined as the point at which the whole mass of a body the point at which the whole mass of a body
may be considered to be concentratedmay be considered to be concentrated. Its coordinate ((xxCMCM, , yyCMCM)) is given the expression below:
xCM=i=1n
mi xi
i=1
n
mi ; y CM=i=1n
mi y i
i=1
n
mi where mi : mass of the i
th particlex i : x coordinate of the i
th particley i : y coordinate of the i
th particle
PHYSICS CHAPTER 8
72
Two masses, 3 kg and 5 kg are located on the yaxis at y =1 m and y =5 m respectively. Determine the centre of mass of this system.Solution :Solution :
Example 8.13 :
0
1 m
y=
5 mm1=3 kg; m2=5 kg
m1
m2
yCM=i=1
2
mi y i
i=1
2
mi
=m1 y1m2 y2
m1m2
yCM=3 1 5 5
35
CM3 . 5 m
PHYSICS CHAPTER 8
73
A system consists of three spheres have the following masses and coordinates :
(1) 1 kg, (3,2) ; (2) 2 kg, (4,5) and (3) 3 kg, (3,0).Determine the coordinate of the centre of mass of the system.Solution :Solution :The x coordinate of the CM is
Example 8.14 :
m1=1 kg; m2=2 kg; m3=3 kg
xCM=i=1
3
mi x i
i=1
3
mi
=m1 x1m2 x2m3 x3
m1m2m3
xCM=1 3 2 4 3 3
123
PHYSICS CHAPTER 8
74
Solution :Solution :The y coordinate of the CM is
Therefore the coordinate of the CM is
yCM=i=1
3
mi y i
i=1
3
y i
=m1 y1m2 y2m3 y3
m1m2m3
yCM=1 2 2 5 3 0
123
PHYSICS CHAPTER 8
75
Figure 8.21 shows a rigid body about a fixed axis O with angular velocity .
is defined as the sum of the products of the mass of each the sum of the products of the mass of each particle and the square of its respective distance from the particle and the square of its respective distance from the rotation axisrotation axis.
8.3.1.2 Moment of inertia, I
m1
m2mn
m3
r1r 2
r3r n
O
Figure 8.21Figure 8.21
PHYSICS CHAPTER 8
76
OR
It is a scalar quantityscalar quantity. Moment of inertia, Moment of inertia, II in the rotational kinematics is analogousanalogous
to the mass, mass, mm in linear kinematics. The dimensiondimension of the moment of inertia is M LM L22. The S.I. unitS.I. unit of moment of inertia is kg mkg m22. The factorsfactors which affect the moment of inertia, I of a rigid body:
a. the massmass of the body,b. the shapeshape of the body,c. the positionposition of the rotation axisrotation axis.
I=m1 r12m2r2
2m3 r32. .. mn rn
2=i=1
n
mi ri2
I : moment of inertia of a rigid body about rotation axism : mass of particler : distance from the particle to the rotation axis
where
PHYSICS CHAPTER 8
77
Moments of inertia of various bodiesMoments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
I CM=MR2
I CM=12 MR
2
Hoop or ring or thin cylindrical
shell
Solid cylinder or disk
CM
CM
PHYSICS CHAPTER 8
78
CM
Moments of inertia of various bodiesMoments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
I CM=1
12 ML2
Uniform rod or long thin rod with
rotation axis through the
centre of mass.
CM
I CM=25 MR
2Solid Sphere
PHYSICS CHAPTER 8
79
Moments of inertia of various bodiesMoments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
I CM=23 MR
2Hollow Sphere or thin spherical
shellCM
Table 8.3Table 8.3
PHYSICS CHAPTER 8
80
Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 8.22.
The spheres are connected by light rods . Determine the moment of inertia of the system about an axisa. through point O,b. along the line AB.
Example 8.15 :
250 cm
60 cm
60 cm
2 kg 3 kg
4 kg5 kg
OA B
Figure 8.22Figure 8.22
PHYSICS CHAPTER 8
81
Solution :Solution :a. rotation axis about point O,
Since r1= r2= r3= r4= r thus and the connecting rods are light therefore
m1=2 kg; m2=3 kg; m3=4 kg; m4=5 kg
r1 0 . 6 m
m1 m2
m3m4
O
r 2
r 4 r31. 25 m
r=0 . 6 21 .25 2=1 .39 mI O=m1 r1
2m2 r22m3 r3
2m4 r 42
I O=r2 m1m2m3m4 =1 .39
2 2345
PHYSICS CHAPTER 8
82
Solution :Solution :b. rotation axis along the line AB,
r1= r2= r3= r4= r=0.6 m therefore
m1=2 kg; m2=3 kg; m3=4 kg; m4=5 kg
I AB=m1 r12m2 r 2
2m3 r32m4 r4
2
I AB=0 . 6 2 2345
m1 m2
m3m4
A B
r1 r 2
r 4 r3
I AB=r2 m1m2m3m4
PHYSICS CHAPTER 8
83
Relationship between torque,Relationship between torque, and angular acceleration, and angular acceleration, Consider a force, F acts on a rigid body freely pivoted on an
axis through point O as shown in Figure 8.23.
The body rotates in the anticlockwise direction and a nett torque is produced.
8.3.2 Torque,
m1
m2
mn r1
r 2
r nO
a1
ana2
F
Figure 8.23Figure 8.23
PHYSICS CHAPTER 8
84
A particle of mass, m1 of distance r1 from the rotation axis O will experience a nett force F1 . The nett force on this particle is
The torque on the mass m1 is
The total (nett) torque on the rigid body is given by
F1=m1 a1F1=m1 r1
a1=r1 and
1=r1 F1 sin 90
1=m1 r12
=i=1n
mi ri2
=m1r12 m2 r22.. .mn r n2
=Ii=1
n
mi ri2=Iand
PHYSICS CHAPTER 8
85
From the equation, the nett torquenett torque acting on the rigid body is proportionalproportional to the bodys angular accelerationangular acceleration.
Note :
Nett torque , =I
Nett force, F=mais analogous to the
PHYSICS CHAPTER 8
86
Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 8.24.
Calculate,a. the nett torque on the disc.b. the magnitude of angular acceleration influence by the disc.( Use the moment of inertia, )
Example 8.16 :
Figure 8.24Figure 8.24
I CM=12 MR
2
F1
O30 .0 cm
F2
PHYSICS CHAPTER 8
87
Solution :Solution :a. The nett torque on the disc is
b. By applying the relationship between torque and angular acceleration,
R=0 . 30 m ; M=5 . 00 kg
=1 2 =RF1RF2=R F1F2 = 0 . 30 5 .6010 . 3
=12 MR2 =I1. 41=12 5 .00 0 . 30 2
PHYSICS CHAPTER 8
88
A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is0.050 kg m2. A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure 8.25. Assume the string does not slip on the wheel.
a. Sketch a free body diagram of the wheel and the block.b. Calculate the magnitude of the angular acceleration of the wheel.
Example 8.17 :
Figure 8.25Figure 8.25
PHYSICS CHAPTER 8
89
Solution :Solution :a. Free body diagram :
for wheel,
for block,
R=0 . 20 m ; I=0 .050 kg m 2 ; P=3 .0 N; m=2 .0 kg
W
TS
NT
W b
P
a
PHYSICS CHAPTER 8
90
Solution :Solution :b. For wheel,
For block,
By substituting eq. (1) into eq. (2), thus
=IRT=I T= IR
(1)
F=ma PT=ma (2)
R=0 . 20 m ; I=0 .050 kg m 2 ; P=3 .0 N; m=2 .0 kg
a=RP IR =ma andP IR =mR
PHYSICS CHAPTER 8
91
An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 8.26. The pulley has a moment of inertia 0.020 kg m2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determinea. the linear acceleration of the object,b. the angular acceleration of the pulley,c. the tension in the string,d. the liner velocity of the object,e. the distance travelled by the object. (Given g = 9.81 m s2)
Example 8.18 :
Figure 8.26Figure 8.26
R
1.50 kg
PHYSICS CHAPTER 8
92
Solution :Solution :a. Free body diagram :
for pulley,
for block,
W
a
T
S =IRT=I = aRandRT=I aR
T= IaR2
(1)T
m g
F=mamgT=ma (2)
PHYSICS CHAPTER 8
93
Solution :Solution :
a. By substituting eq. (1) into eq. (2), thus
b. By using the relationship between a and , hence
mg IaR2 =maR=0 .20 m ; I=0 . 020 kg m 2 ; m=1 . 50 kg;u=0; t=0 .3 s
a=R
1 .50 9 .81 0 .020 a0 . 20 2 =1 .50 a7 . 36=0 . 20
PHYSICS CHAPTER 8
94
Solution :Solution :
c. From eq. (1), thus
d. By applying the equation of liner motion, thus
e. The distance travelled by the object in 0.3 s is
R=0 .20 m ; I=0 . 020 kg m 2 ; m=1 . 50 kg;u=0; t=0 .3 s
v=uatv=0 7 .36 0 . 3
T= IaR2
T= 0 . 020 7 . 36 0 .20 2
s=ut12 at2
s=0127 . 36 0 .3 2
PHYSICS CHAPTER 8
95
Exercise 8.4 :Use gravitational acceleration, g = 9.81 m s21. Three oddshaped blocks of chocolate have following masses
and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m); 0.400 kg, (0.100 m. 0.400 m); 0.200 kg, (0.300 m, 0.600 m).Determine the coordinates of the centre of mass of the system of three chocolate blocks.
ANS. : ANS. : (0.044 m, 0.056 m)(0.044 m, 0.056 m)2. Figure 8.27 shows four masses that are held at
the corners of a square by a very light frame. Calculate the moment of inertia of the system about an axis perpendicular to the planea. through point A, andb. through point B.
ANS. : ANS. : 0.141 kg m0.141 kg m22; 0.211 kg m; 0.211 kg m22
80 cm
80 cm
150 g 150 g
70 g
70 g
40 cm
A
B
Figure 8.27Figure 8.27
PHYSICS CHAPTER 8
96
2 . 00 m s2
T 2
T 1
Exercise 8.4 :3. A 5.00 kg object placed on a
frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 8.28. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s2.a. Sketch free body diagrams of both objects and pulley.b. Calculate T1 and T2 the tensions in the string.c. Determine I.
ANS. : 10.0 N, 70.3 N; 1.88 kg mANS. : 10.0 N, 70.3 N; 1.88 kg m22
Figure 8.28Figure 8.28
PHYSICS CHAPTER 8
97
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Solve problem related to :Solve problem related to :
kinetic energy, kinetic energy,
work, work,
power, power,
Learning Outcome:8.4 Work and Energy of Rotational Motion (2 hours)
K r=12 I
2
P=
W=
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PHYSICS CHAPTER 8
98
8.4 Rotational kinetic energy and power8.4.1 Rotational kinetic energy, Kr Consider a rigid body rotating about the axis OZ as shown in
Figure 8.29.
Every particle in the body is in the circular motion about point O.
m1
m2
mn
m3
r1r 2
r3
r nO
v1v2
v3vn
Z
Figure 8.29Figure 8.29
PHYSICS CHAPTER 8
99
The rigid body has a rotational kinetic energy which is the total total of kinetic energy of all the particles in the bodyof kinetic energy of all the particles in the body is given by
K r=12 m1 v1
212 m2 v2
212 m3 v 3
2.. .12 mn vn2
K r=12 m1 r1
2212 m2 r 22 212 m3 r3
2 2. .. 12 mn rn2 2
K r=12
2 m1r12m2r 22m3r32. . .mn rn2
K r=12 I
2
K r=12
2i=1n
mi r i2 i=1
n
mi r i2=Iand
PHYSICS CHAPTER 8
100
From the formula for translational kinetic energy, Ktr
After comparing both equations thus
For rolling body without slippingrolling body without slipping, the total kinetic energy of total kinetic energy of the body, the body, KK is given by
K tr=12 mv
2
is analogous to vvII is analogous to mm
K=K trK rK tr : translational kinetic energyK r : rotational kinetic energy
where
PHYSICS CHAPTER 8
101
A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25 to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculatea. the total kinetic energy of the sphere,b. the linear speed of the sphere,c. the angular speed about the centre of mass.(Given the moment of inertia of solid sphere is and the gravitational acceleration, g = 9.81 m s2)
Example 8.19 :
I CM=25 mR
2
PHYSICS CHAPTER 8
102
Solution :Solution :
a. From the principle of conservation of energy,
R=0 .15 m ; m=10 . 0 kg
E i= E fmgh=K
K=mgs sin 25K=10 . 0 9 . 81 0 . 75 sin 25
s=0 . 75 m
h=ssin 25
v CM 25
R
PHYSICS CHAPTER 8
103
Solution :Solution :b. The linear speed of the sphere is given by
c. By using the relationship between v and , thus
R=0 .15 m ; m=10 . 0 kg
K=K trK r K=12 mv
212 I
2 = vRand
K=12
mv212 25 mR2 vR
2
K= 710 mv2
31 .1= 71010 . 0 v2
v=R 2 .11=0 .15
PHYSICS CHAPTER 8
104
The pulley in the Figure 8.30 has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s2)
Example 8.20 :
2 .00 kg
5 . 00 kg
7 . 00 m
Figure 8.30Figure 8.30
PHYSICS CHAPTER 8
105
Solution :Solution :The moment of inertia of the pulley,
m1=5 .00 kg ;m2=2.00 kg; R=0 .120 m; h=7 . 00 m
I= 0 . 055 g 1 cm2103 kg1 g 104m2 1 cm2 =5 . 5109 kg m2
m2
m1
7 . 00 m
Initial
m2
m17 . 00 m
v
v
Final
E i=U1 E f =K tr 1K tr 2K rU2
PHYSICS CHAPTER 8
106
Solution :Solution :
By using the principle of conservation of energy, thus
E i= E fU1=K tr1K tr 2K rU2
m1 gh=12 m1 v
212 m2 v
212 I
2m2 gh
m1m2 gh=12
v2 m1m2 12
I vR 2
5 . 002 . 00 9 . 81 7 . 00=12
v 2 5 . 002. 00 125 . 5109 v0 . 120
2
m1=5 . 00 kg ;m2=2. 00 kg; R=0 .120 m;h=7 . 00 m; I=5 .5109 kg m2
PHYSICS CHAPTER 8
107
Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 8.31.
The work done by the tangential force is given by
8.4.2 Work, W
Figure 8.31Figure 8.31
F
ds
O
dRR
dW=FdsdW=FRd dW=1
2 d W=12 d
ds=Rdand
PHYSICS CHAPTER 8
108
If the torque is constant thus
Workrotational kinetic energy theorem states
W= 21W=1
2 d
W=
: torque : change in angular displacement
where
W : work done
W=K r= K r f K r iW=12 I
212 I0
2
is analogous to the W=Fs
PHYSICS CHAPTER 8
109
From the definition of instantaneous power,
Caution : The unitunit of kinetic energy, work and powerkinetic energy, work and power in the
rotationalrotational kinematics is samesame as their unitunit in translational translational kinematics.
8.4.3 Power, P
P=dWdt dW=dand
P= ddtP=
ddt =
and
is analogous to the P=Fv
PHYSICS CHAPTER 8
110
A horizontal merrygoround has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merrygo round for 15.0 s. If the merrygoround is initially at rest and ignore the frictional torque, determinea. the rotational kinetic energy of the merrygoround,b. the work done by the force on the merrygoround,c. the average power supplied by the force.(Given g = 9.81 m s2)Solution :Solution :
Example 8.21 :
FR=2 . 40 m
PHYSICS CHAPTER 8
111
Solution :Solution :
a. By applying the relationship between nett torque and angular acceleration, thus
Use the equation of rotational motion with uniform angular acceleration,
Therefore the rotational kinetic energy for 15.0 s is
=IRF=I 2 . 40 18 . 0 =2100
=0t=0 2 .06102 15 .0 =0 .309 rad s1
K r=12 I
2 K r=122100 0 .309 2
R=2 . 40 m ; I=2100 kg m2 ; F=18 . 0 N;t=15 .0 s; 0=0
PHYSICS CHAPTER 8
112
Solution :Solution :
b. The angular displacement, for 15.0 s is given by
By applying the formulae of work done in rotational motion, thus
c. The average power supplied by the force is
W=
=0 t12 t
2
Pav=Wt
W=2 . 40 18 . 0 2 . 32
R=2 . 40 m ; I=2100 kg m2 ; F=18 . 0 N;t=15 .0 s; 0=0
=0122. 0610215 . 0 2
W=RF
Pav=10015 . 0
PHYSICS CHAPTER 8
113
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use angular momentum, angular momentum,
State and useState and use the principle of conservation of angular the principle of conservation of angular momentummomentum
Learning Outcome:
8.5 Conservation of angular momentum (1 hour)
L=I
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PHYSICS CHAPTER 8
114
8.5 Conservation of angular momentum8.5.1 Angular momentum, is defined as the product of the angular velocity of a body the product of the angular velocity of a body
and its moment of inertia about the rotation axisand its moment of inertia about the rotation axis. OR
It is a vectorIt is a vector quantity. Its dimension is M LM L22 T T11 The S.I. unit of the angular momentum is kg mkg m22 s s11.
L
where
L=IL : angular momentumI : moment of inertia of a body : angular velocity of a body
is analogous to the p=mv
PHYSICS CHAPTER 8
115
The relationship between angular momentum, L with linear momentum, p is given by vector notation :
magnitude form :
Newtons second law of motion in term of linear momentum is
hence we can write the Newtons second law in angular form as
and states that a vector sum of all the torques acting on a a vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular rigid body is proportional to the rate of change of angular momentummomentum.
L=rp=rm v
L=rpsin =mvr sin where
:the angle between {r with {v r : distance from the particle to the rotation axis
F=Fnett=d pdt
= nett=dL
dt
PHYSICS CHAPTER 8
116
states that a total angular momentum of a system about an a total angular momentum of a system about an rotation axis is constant if no external torque acts on the rotation axis is constant if no external torque acts on the systemsystem.OR
Therefore
8.5.2 Principle of conservation of angular momentum
I=constant
=d Ldt =0d L=0
If the =0
dL= L f Li L i= L f
and
PHYSICS CHAPTER 8
117
A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s1 about the rotation axis as shown in Figure 8.32 . A 50 kg bag of sand falls onto the disc at the edge of the wooden disc.
Calculate,a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle)b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same?(Use the moment of inertia of disc is )
Example 8.22 :
0
Before
R
After
R
Figure 8.32Figure 8.32
12 MR
2
PHYSICS CHAPTER 8
118
Solution :Solution :a. The moment of inertia of the disc,
The moment of inertia of the bag of sand,
By applying the principle of conservation of angular momentum,
R=3 . 00 m ;0=4 . 0 rad s1 ; mw=200 kg; mb=50 kg
I w=12 mw R
2=12200 3. 00 2
I w 0=I wI b
I b=mb R2= 50 3 . 002
L i= L f
I w=900 kg m2
I b=450 kg m2
900 4 .0 = 900450
PHYSICS CHAPTER 8
119
Solution :Solution :b. The initial rotational kinetic energy,
The final rotational kinetic energy,
thus It is because the energy is lost in the form of heat from the the energy is lost in the form of heat from the friction between the surface of the disc with the bag of friction between the surface of the disc with the bag of sand.sand.
R=3 . 00 m ;0=4 . 0 rad s1 ; mw=200 kg; mb=50 kg
K r i=12 I w 0
2=12900 4 .0 2
K r f=12 I wI b
2=12900450 2 . 672
K r i=7200 J
K r iK r f
PHYSICS CHAPTER 8
120
A raw egg and a hardboiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer.Solution :Solution :The answer is hardboiled egghardboiled egg.
Example 8.23 :
PHYSICS CHAPTER 8
121
Solution :Solution :ReasonReasonRaw egg : When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases because ofFrom the principle of conservation of angular momentum,
If the I is increases hence its angular velocity, will decreases.
Hardboiled egg :The position of the yolk of a hardboiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.
I=mr2
I=constant
PHYSICS CHAPTER 8
122
A student on a stool rotates freely with an angular speed of 2.95 rev s1. The student holds a 1.25 kg mass in each outstretched arm that is 0.759 m from the rotation axis. The moment of inertia for the system of studentstool without the masses is 5.43 kg m2. When the student pulls his arms inward, the angular speed increases to 3.54 rev s1. a. Determine the new distance of each mass from the rotation axis.b. Calculate the initial and the final rotational kinetic energy of the system.Solution :Solution :
Example 8.24 :
0=2 .95 rev1 s 2 rad1 rev =18 . 5 rad s1
=3 . 54 rev1 s 2 rad1 rev =22. 2 rad s1
PHYSICS CHAPTER 8
123Before
0
After
r a r a
Solution :Solution :
r br bmm
m=1. 25 kg ; 0=18. 5 rad s1 ; I ss=5. 43 kg m
2 ;r b=0 .759 m ; =22. 2 rad s
1 ;
PHYSICS CHAPTER 8
124
Solution :Solution :
a. The moment of inertia of the system initially is
The moment of inertia of the system finally is
By using the principle of conservation of angular momentum, thus
I i=I ssI m I i=I ssmrb2mrb2 =I ss2 mrb2I i=5 .43 2 1 .25 0 .759
2=6 . 87 kg m2
I f=I ss2 mr a2= 5 .43 2 1.25 ra2
I i 0=I f Li= Lf
6 . 87 18 .5 =5 .432 .5r a222 .2
m=1. 25 kg ; 0=18. 5 rad s1 ; I ss=5. 43 kg m
2 ;r b=0 .759 m ; =22.2 rad s
1 ;
I f =5 . 432 .5r a2
PHYSICS CHAPTER 8
125
Solution :Solution :
b. The initial rotational kinetic energy is given by
and the final rotational kinetic energy is
K r i=12 I i 02
=126 .87 18 . 52
K r i=1 .18103 J
K r f=12 I f
2
=125 . 432 .5 0 .344 222. 2 2
m=1. 25 kg ; 0=18. 5 rad s1 ; I ss=5. 43 kg m
2 ;r b=0 .759 m ; =22.2 rad s
1 ;
PHYSICS CHAPTER 8
126
Exercise 8.5 :Use gravitational acceleration, g = 9.81 m s21. A woman of mass 60 kg stands at the rim of a horizontal
turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s1 relative to the Earth.a. In the what direction and with what value of angular speed
does the turntable rotate?b. How much work does the woman do to set herself and the
turntable into motion?ANS. : ANS. : 0.360 rad s0.360 rad s11 ,U think; 99.9 J ,U think; 99.9 J
PHYSICS CHAPTER 8
127
Exercise 8.5 :2. Determine the angular momentum of the Earth
a. about its rotation axis (assume the Earth is a uniform solid sphere), andb. about its orbit around the Sun (treat the Earth as a particle
orbiting the Sun). Given the Earths mass = 6.0 x 1024 kg, radius = 6.4 x 106 m
and is 1.5 x 108 km from the Sun.ANS. : ANS. : 7.1 x 107.1 x 103333 kg m kg m22 s s11; 2.7 x 10; 2.7 x 104040 kg m kg m22 s s11
3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is )
ANS. : ANS. : 4.71 x 104.71 x 1066 kg m kg m22 s s11
112 ML
2
PHYSICS CHAPTER 8
128
Linear Motion Relationship Rotational MotionSummary:
m
v=r =ddt
a=dvdt =ddt
F=ma =I
W=Fs W=
v=dsdt
a=r
=rF sin
P=Fv P=
p=mv L=IL=rpsin
II=i=1
n
mi ri2
129
PHYSICS CHAPTER 8
THE ENDNext Chapter
CHAPTER 9 :Simple Harmonic Motion (SHM)
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Slide 1Kinematics of linear motion Learning Outcome:2.1. Linear motion (1D)Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 262.2. Uniformly accelerated motionSlide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 402.3. Free falling bodySlide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 532.4. Projectile motionSlide 55Slide 56Slide 57Slide 58Slide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Slide 68Slide 69Slide 70Slide 71Slide 72Slide 73Slide 74Slide 75Slide 76Slide 77Slide 78Slide 79Slide 1Slide 2Learning Outcome:Slide 43.1.2 Impulse, Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 133.2 Conservation of linear momentum Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 223.2.2 CollisionSlide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 1Learning Outcome:7.1 Parameters in rotational motionSlide 47.1.2 Angular velocitySlide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 137.1.3 Angular accelerationSlide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 237.3 Rotational motion with uniform angular acceleration Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 405.3.2 Torque (moment of a force), Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 495.3.3 Equilibrium of a rigid bodySlide 51Slide 525.3.4 Problem solving strategies for equilibrium of a rigid bodySlide 54Slide 55Slide 56Slide 57Slide 58Slide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Slide 68Slide 69Slide 707.4 Centre of mass, moment of inertia and torqueSlide 72Slide 73Slide 747.4.2 Moment of inertia, ISlide 76Slide 77Slide 78Slide 79Slide 80Slide 81Slide 827.4.3 Torque,Slide 84Slide 85Slide 86Slide 87Slide 88Slide 89Slide 90Slide 91Slide 92Slide 93Slide 94Slide 95Slide 96Slide 977.5 Rotational kinetic energy and powerSlide 99Slide 100Slide 101Slide 102Slide 103Slide 104Slide 105Slide 1067.5.2 Work, WSlide 1087.5.3 Power, PSlide 110Slide 111Slide 112Slide 1137.6 Conservation of angular momentumSlide 1157.6.2 Principle of conservation of angular momentumSlide 117Slide 118Slide 119Slide 120Slide 121Slide 122Slide 123Slide 124Slide 125Slide 126Slide 127Summary:Slide 129
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