chapter 8 matriculation stpm

7/28/2019 Chapter 8 Matriculation STPM

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PHYSICS CHAPTER 8

CHAPTER 8:CHAPTER 8:Rotational of rigid bodyRotational of rigid body

(8 Hours)(8 Hours)

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PHYSICS CHAPTER 8

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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:

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) Relateb) Relate parameters in rotational motion with their correspondingparameters in rotational motion with their corresponding

quantities in linear motion.quantities in linear motion. Write and useWrite and use ::

c)c) UseUse equations for rotational motion with constant angularequations for rotational motion with constant angular

acceleration.acceleration.

Learning Outcome:

8.1 Rotational Kinematics (2 hour)

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s=r;v=r;at=r;a

c=r

2=

v2

r

=0t =0 t1

2 t2

2

=022


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PHYSICS CHAPTER 8

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8.1 Parameters in rotational motion

8.1.1 Angular displacement, is defined as an angle through which a point or line hasan 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 Pon a rotating compact disc (CD)moves through an arc lengths on a circular path of radius rabout a fixed axis through point O.

Figure 8.1Figure 8.1


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PHYSICS CHAPTER 8

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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 angularthe instantaneous rate of change of angular

displacementdisplacement.

Equation :

8.1.2 Angular velocity

av=

2

1

t2t1=

t

= limitt 0

t=d

dt

where 2: final angular displacement in radian

t: time interval

1: initial angular displacement in radian


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PHYSICS CHAPTER 8

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It is a vector quantityvector quantity.

The unit of angular velocity is radian per second (rad sradian per second (rad s-1-1))

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 angularsame 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 =2

60

rad s1=

30

rad s1

ThumbThumb : direction ofangular velocityangular velocity

Curl fingersCurl fingers : direction ofrotationrotation


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PHYSICS CHAPTER 8

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Figures 8.2 and 8.3 show the right hand grip rule for determining

the direction of the angular velocity.




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PHYSICS CHAPTER 8

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The angular displacement, of the wheel is given by

where in radians and tin seconds. The diameter of the wheel is0.56 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 :

=5t2t


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PHYSICS CHAPTER 8

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Solution :Solution :

a. At time, t1=2.2 s :

At time, t2=4.8 s :

r=d

2

=0.56

2

=0.28 m

1=5 2. 2

2 2. 2

1=22 rad

2=5 4 . 8

24 . 8

2=110 rad


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PHYSICS CHAPTER 8

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b. By applying the equation of arc length,

Therefore

c. The average angular velocity in rad s1 is given by

r=d

2

=0.56

2

=0.28 m

s=rs=r=r21

s=0.28 11022

av=

t

=21

t2t1

av=

110224 .82.2


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PHYSICS CHAPTER 8

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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 rad

1 s 1 rev

2 rad 60 s

1 min

=d

dt

5t2t

=d

dt

=10 t1

=10 3 .0 1


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PHYSICS CHAPTER 8

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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.

(Giveng= 9.81 m s2)


Example 8.2 :

uy=0

0=0

10 m

water

1=2. 5 rev


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PHYSICS CHAPTER 8

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From the diagram,

Thus

Therefore the divers average angular velocity is

1=2. 52=5 rad

sy

=10 m

sy=uy t1

2gt2

10=01

29.81 t2

av=10

t

av=501.43


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PHYSICS CHAPTER 8

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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 angularthe instantaneous rate of change of angular

velocityvelocity.

Equation :

8.1.3 Angular acceleration

av=

2

1

t2t1=

t

= limitt0

t=d

dt

where 2 : final angular velocity

t: time interval

1: initial angular velocity


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PHYSICS CHAPTER 8

15Figure 8.4Figure 8.4


The unit of angular acceleration is rad srad s22. Note:

If the angular acceleration, is positivepositive, then the angularvelocity, is increasingincreasing.

If the angular acceleration, is negativenegative, then the angularvelocity, is decreasingdecreasing.

Direction of the angular accelerationDirection of the angular acceleration

If the rotation is speeding upspeeding up, and in the same directionsame directionas shown in Figure 8.4.


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PHYSICS CHAPTER 8

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If the rotation is slowing downslowing down, and have the oppositeoppositedirectiondirection 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 tin seconds.

Determine

a. the average angular acceleration between 2.2 s and 4.8 s,

b. the instantaneous angular acceleration at time, 3.0 s.

=8t3

t2


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PHYSICS CHAPTER 8

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a. At time, t1=2.2 s :

At time, t2=4.8 s :

Therefore the average angular acceleration is

1=8 2 . 2 32 . 2 2

1=80 .3 rad s1

2=8 4 . 8

34 . 8

2

av=

2

1

t2t1

av=86280.3

4.82.2


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PHYSICS CHAPTER 8

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b. The instantaneous angular acceleration as a function of time is

At time, t=3.0 s :

=d

dt8t3t2

= ddt

=24 3.0 22 3.0


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PHYSICS CHAPTER 8

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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; 138

radrad2. During a certain period of time, the angular displacement of a

swinging door is described by

where is in radians and tis in seconds. Determine the

angular displacement, angular speed and angular acceleration

a. 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


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PHYSICS CHAPTER 8

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8.1.2 Relationship between linear and

rotational motion8.1.2 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

r

P

O



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PHYSICS CHAPTER 8

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Point P moves in a circle of radius rwith 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, rris changingchangingdependdepend on the position of the particleposition of the particle.

v=ds

dt

v=rd

dt

s=r

v=r

and

Simulation 7.1
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PHYSICS CHAPTER 8

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at

ac

a

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,

atand angular acceleration,



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PHYSICS CHAPTER 8

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The components are tangential acceleration,tangential acceleration, aattand

centripetal acceleration,centripetal acceleration, aaccgiven 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, aagiven by

and its magnitude,

at=dv

dt

a t=rd

dt

at=r

v=rand

ac=v

2

r=r2=v

a=ata

c

a=at2ac2Vector formVector form


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PHYSICS CHAPTER 8

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8.1.3 Rotational motion with uniform

angular acceleration Table 8.1 shows the symbols used in linear and rotationalkinematics.

Table 8.1Table 8.1

Linearmotion

QuantityRotational

motion

s DisplacementDisplacement

u 0Initial velocityInitial velocity

v Final velocityFinal velocity

a AccelerationAcceleration

t tTimeTime


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PHYSICS CHAPTER 8

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Table 8.2 shows the comparison of linear and rotational motion

with constant acceleration.

Linear motion Rotational motion

a=constant

v=uat

=constant

=0

t

s=ut1

2at2

=0 t1

2t2

v

2

=u2

2as 2

=02

2

s=1

2vu t =

1

20 t

where in radian. Table 8.2Table 8.2


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PHYSICS CHAPTER 8

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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, calculate

a. 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 s

u=r0

17.0=0.480

2.00=0.48a=r


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PHYSICS CHAPTER 8

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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 t1

2t2

=

229 rad

u=17 . 0 m s1

, r=0 . 48 m, a=2 . 00 m s2

, t=5 . 00 s

=35.4 5.001

24.17 5.002

=0t=35 .44.17 5.00


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PHYSICS CHAPTER 8

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The wheels of a bicycle make 30 revolutions as the bicycle

reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. Thewheels 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 km

1 h 103m

1 km 1 h3600 s =13. 9 m s1,v=

35.0 km

1 h 103m

1 km 1 h3600 s =9.72 m s1


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PHYSICS CHAPTER 8

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a. The initial angular speed of the wheels is

and the final angular speed of the wheels is

therefore

b. The car stops thusHence

u=r013.9=0.350

v=r

9.72=0.352=0

22

27.8 2=39.722 60

=0 0=27.8 rad s1

and=

0t

0=27.8 2.13 t


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PHYSICS CHAPTER 8

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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 s-1. Theangular acceleration of the blade is 0.750 rev s -2. Determine

a. 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.


a. Given t=4.00 s, thus

Example 8.6 :

r=0 .400 m,0=0.1502=0.300 rad s

1,

=0.7502=1.50 rad s2

=0t

=19. 8 rad s1

=0.3001.50 4.00


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PHYSICS CHAPTER 8

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b. The number of revolutions of the blade is

c. The tangential speed of a point is given by

=0 t12t2

=41.5 rad

=0.300 4.001

21.50 4.00

2

v=r

v=0.400 19.8


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PHYSICS CHAPTER 8

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d. The magnitude of the resultant acceleration is

a=ac2at2

a=

v 2

r

2

r 2

a=7.922

0.400 2

0.4001.50 2


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PHYSICS CHAPTER 8

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Calculate the angular velocity of

a. the second-hand,

b. the minute-hand and

c. the hour-hand,

of a clock. State in rad s-1.

d. What is the angular acceleration in each case?Solution :Solution :

a. The period of second-hand of the clock is T= 60 s, hence

Example 8.7 :

=

2

T =

2

60


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PHYSICS CHAPTER 8

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b. The period of minute-hand of the clock is T= 60 min = 3600 s,

hence

c. The period of hour-hand of the clock is T= 12 h = 4.32 104 s,

hence

d. The angular acceleration in each cases is

=2

3600

=2

4.3210

4


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PHYSICS CHAPTER 8

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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 speedof 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.


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=d

2=1.2010

2m


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PHYSICS CHAPTER 8

36


The initial speed of the point at the edge the coin is

and the final speed is

The 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 s

1

a=r

a=1.20102 1.90

v2=u22as0=0.216

22 2.28102 s

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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Exercise 8.2 :

1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev

min-1 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 101033 m sm s22; 20.1 m; 20.1 m

2. A 0.35 m diameter grinding wheel rotates at 2500 rpm.

Calculate

a. 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 101044 m sm s22

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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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 is98.0 rad s-1. 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 of3.50 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


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PHYSICS CHAPTER 8

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Exercise 8.2 :

5. A bicycle wheel is being tested at a repair shop. The angular

velocity of the wheel is 4.00 rad s-1 at time t= 0 , and its angularacceleration is constant and equal 1.20 rad s-2. A spoke OP on

the wheel coincides with the +x-axis 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 x-axis at this time?

ANS. :ANS. : 0.40 rad s0.40 rad s11; 18; 18


x

y

P

O

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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Define and useDefine and use torque.torque.

State and useState 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


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PHYSICS CHAPTER 8

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8.2.1 Torque (moment of a force),

The magnitude of the torquemagnitude of the torque is defined as the product of athe product of a

force and its perpendicular distance from the line of actionforce 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 of

where r: distance between the pivot point (rotationaxis) and the point of application of force.

Thus

Fd=

forcetheofmagnitude:Farm)(momentdistancelarperpendicu:d

torquetheofmagnitude:where

sinrd =

sin Fr=

rF

andbetweenangle:

where

OR Fr

=

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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The dimension of torque is

The unit of torqueunit of torque is N mN m (newton metre), a vector productvector product

unlike thejoule (unit of work)joule (unit of work), also equal to a newton metre,

which is scalar productscalar product.

Torque is occurred because ofturning (twisting) effects ofturning (twisting) effects of

the forcesthe 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 oftorque dependstorque depends on the rotation axisrotation axis and the

magnitude of applied forcemagnitude of applied force.

[ ] [ ][ ] 22TMLdF ==

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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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


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PHYSICS CHAPTER 8

44

O


2

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 1

r

0sin=== 333333

rFdF

222 rd sin=

111111 rFdF sin==222222 rFdF sin==

2r

2211 dFdF =O

Fr sin=

0=

and

0=

Simulation 5.1

PHYSICS CHAPTER 8
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PHYSICS CHAPTER 8

45

Determine a resultant torque of all the forces about rotation axis, O

in the following problems.a.

Example 8.9 :

m5

N10=2F

m5 N30=1F

m3

m3

N20=3F

m10

m6O

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

46

b.

Example 8.9 :

m5

N10=2F

m5

N30=

1F

m3

m3

N25=4F

N20=3F

m10

m6O

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

47

m5m5

m10

m6O


a.

Force Torque (N m), o=Fd=Frsin

1F

( ) ( ) 90330 =

2F ( ) ( ) 50510 +=+

N10=2F

N30=1F

N20=3

F

m3=1d

m5=2d

3F

0The resultant torque:

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

48

m5

m10

m3

m6

m5


b.

Force Torque (N m), o=Fd=Frsin

1F

( ) ( ) 90330 =

2F

( )( ) ( ) 51.50.515520sin ==rF33F

0 The resultant torque:

N10=2F

N30=1F

0.515

53

3sin

22=

+

=

O

N20=3F

N25=4F

m3=1d

m5=r

4F

0

3d

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

49

8.2 Equilibrium of a rigid body

8.2.1.1 Non-concurrent forces is defined as the forces whose lines of action do not passthe 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 non-concurrent forces cause

rolling motionrolling motion on the body. (translational and rotationaltranslational and rotationalmotion)

Figure 8.11 shows an example of non-concurrent forces.

2F

3F

1F


4F

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

50

8.2.1.2 Equilibrium of a rigid body

Rigid bodyRigid body is defined as a body with definite shape thata body with definite shape that

doesnt change, so that the particles that compose it stay indoesnt change, so that the particles that compose it stay in

fixed position relative to one another even though a force isfixed 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 mustThe vector sum of all forces acting on a rigid body must

be zero.be zero.

==0nettFF

OR

=== 0,0,0 zyx FFF

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

51

The vector sum of all external torques acting on a rigidThe 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, CGCentre of gravity, CG is defined as the point at which the whole weight of a bodythe 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


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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.

2

l

2

l

CG

CGl



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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

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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, calculate

a. the weight, W1 needed,

b. the force exerted on the beam at point O.

(Giveng=9.81 m s2)

Example 8.10 :

1W2W

A BO35 cm 75 cm


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The free body diagram of the beam :

Let point O as the rotation axis.

N23;kg3 == 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 y-comp. (N) Torque (N m), o=Fd=Frsin

1W

1W

gm ( ) ( )9.813 ( ) ( ) 5.880.2029.4 =

( ) 11 WW 0.750.75 =

2W

23 ( ) ( ) 8.050.3523 =+

N

N 0

29.4=

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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

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A uniform ladder AB of length 10 m and

mass 5.0 kg leans against a smooth wallas 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 mup the ladder , Calculate the coefficient

of static friction between ladder and

floor.

(Giveng=9.81 m s2)

Example 8.11 :

A

B

smooth

wall

rough floor


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a. The free body diagram of the ladder :

Let point B as the rotation axis.

kg60;kg5.0 == fl mm

A

B

CG

gmf

1N

gml

2N

m8.0m10

m3.0

m5.0

Forcex-comp.

(N)y-comp.

(N)

Torque (N m),

B=Fd=Frsin

gml

1N1N

0.810

8sin ==

sf

gmf

49.1 0.6

10

6sin ==

2N

sf

0

5890

2N

0

0

m6.0

( ) ( ) sin5.049.1

147=

0

( ) ( ) sin3.05891060=

( ) N1 sin101N8=

0

0 sf

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Since the ladder in equilibrium thus

0= B081060147 =+ 1N

N151=1N

0= xF 0= s1 fNHorizontal force:Horizontal force:

0=yF

058949.1 =+ 2NVertical force:Vertical force:

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60

m10

A

B

m8.0

m6.0

m5.0


b. The free body diagram of the ladder :

Let point B as the rotation axis.

0.6sin0.8;sin ==

gmf

gml

sf

m7.0

Forcex-comp.

(N)y-comp.

(N)

Torque (N m),

B=Fd=Frsin

gml

1N1N

2sN

gmf

49.1

2N

sf

0

5890

2N

0

0

( ) ( ) sin5.049.1

147=

0

( ) ( ) sin7.05892474=

( ) N1 sin101N8=

0

0

2N

1N

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Consider the ladder stills in equilibrium thus

0= B082474147 =+ 1N

N328=1N

0=

x

F

0= 2s1 NN

0= yF 058949.1 =+ 2NN638=2N

( ) ( ) 0638328 = s

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A floodlight of mass 20.0 kg in a park is

supported at the end of a 10.0 kg uniformhorizontal 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.

(Giveng=9.81 m s2

)

Example 8.12 :

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a. The free body diagram of the beam :

b. Let point O as the rotation axis.

kg10.0;kg20.0 == bf mm

Force x-comp. (N) y-comp. (N) Torque (N m), o=Fd=Frsin

gmf

( )l1960 196

O CG

gmf

gmb

T

S

30

l

l0.5

gmb ( ) ( ) ll 49.10.598.1 =0 98.1

T

TlTl 0.530sin =30cosT 30sinT

S

xS yS 0

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b. The floodlight and beam remain at rest thus

i.

ii.

0= O 00.549.1196 =+ Tlll

0=xF

0cos =+ xS30T

N424=xS0=

y

F

030sin98.1196 =++ yST

N49.1=yS

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b. ii. Therefore the magnitude of the force is

and its direction is given by

2y

2x SSS +=

( ) ( ) 22S 49.1424 +=

=

x

y

S

S 1tan

=

42449.1tan 1

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Exercise 8.3 :

Use gravitational acceleration,g= 9.81 m s2

1.

Figure 8.17 shows the forces,F1 =10 N,F2= 50 N andF3=

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 resultanttorque about point D.

ANS. : -3.7 N mANS. : -3.7 N m

D

AB

C

1F

3F

2F


a

b

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Exercise 8.3 :

2.

A see-saw 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. Determine

a. 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

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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 8.0kg. 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


Exercise 8.3 :

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4.

A uniform 10.0 N picture frame is supported by two light string

as shown in Figure 8.20. The horizontal force,Fis applied forholding 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 :


F

50.0

cm15.0

cm30.0

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Learning Outcome:

8.3 Rotational dynamics (1 hour)

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DefineDefine 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

1i

iirmI2

I =

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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 bodythe 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

mixi

i=1

n

mi ; y CM=i=1

n

m i y i

i=1

n

mi

where mi : mass of the ith

particlexi :x coordinate of the i

thparticle

y i : y coordinate of the ith

particle

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Two masses, 3 kg and 5 kg are located on the y-axis 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

=

5 mm

1=3 kg; m

2=5 kg

m1

m2

yCM=

i=1

2

miyi

i=1

2

mi

=m

1y

1m

2y

2

m1m 2

yCM=

3 1 5 5

35

CM3. 5 m

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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.


Thex coordinate of the CM is

Example 8.14 :

m1=1 kg; m

2=2 kg; m

3=3 kg

xCM=

i=1

3

mi x i

i=1

3

mi

=m

1x

1m

2x

2m

3x

3

m1m

2m

3

xCM=

1 3 2 4 3 3

123

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They coordinate of the CM is

Therefore the coordinate of the CM is

yCM=

i=1

3

miyi

i=1

3

yi

=m

1y

1m

2y

2m

3y

3

m1m

2m

3

yCM=

1 2 2 5 3 0

123

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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 eachthe sum of the products of the mass of each

particle and the square of its respective distance from theparticle and the square of its respective distance from the

rotation axisrotation axis.

8.3.1.2 Moment of inertia,I

m1

m2

mn

m3

r1

r2r3

rn O


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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,Iof a rigid body:a. the massmass of the body,

b. the shapeshape of the body,

c. the positionposition of the rotation axisrotation axis.

I=m1r

12m

2r

22m

3r

32. ..m

nrn2=

i=1

n

miri2

I : moment of inertia of a rigid body about rotation axism : mass of particler : distance from the particle to the rotation axis

where

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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

ICM=MR 2

ICM=1

2MR

2

Hoop or ring or

thin cylindrical

shell

Solid cylinder or

diskCM

CM

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CM





ICM=1

12 ML2

Uniform rod or

long thin rod with

rotation axis

through the

centre of mass.

CM

ICM=2

5MR

2Solid Sphere

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ICM=23MR2Hollow Sphere orthin spherical

shell

CM

Table 8.3Table 8.3

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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 momentof inertia of the system about an axis

a. 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


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a. rotation axis about point O,

Since r1= r

2= r

3= r

4= r thus

and the connecting rods are light therefore

m1=2 kg; m

2=3 kg; m

3=4 kg; m

4=5 kg

r1

0. 6 m

m1

m2

m3m4

O

r2

r4

r3

1.25 m

r=0 . 6 21.25 2=1 .39 mIO=m1r12m2r22m3r32m4r42

IO=r

2 m1m2m3m4 =1.39 2 2345

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b. rotation axis along the line AB,

r1= r

2= r

3= r

4= r=0.6 m therefore

m1=2 kg; m

2=3 kg; m

3=4 kg; m

4=5 kg

IAB=m1r12m2r2

2m3r3

2m4 r4

2

IAB=0 . 6

22345

m1

m2

m3m4

A B

r1

r2

r4 r3

IAB=r2

m1m2m3m4

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Relationship between torque,Relationship between torque, and angular acceleration,and angular acceleration, Consider a force,Facts 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

r2

rn

O

a1

an

a2

F


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A particle of mass, m1

of distance r1

from the rotation axis O will

experience a nett forceF1

. 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=m1a1F

1=m

1r

1

a1=r1 and

1=r

1F

1sin90

1=m1r12

=i=1n

mir

i

2

=m1r1

2m2 r2

2.. .mnrn

2

=I

i=1

n

m i ri2=Iand

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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

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Forces,F1 = 5.60 N andF2= 10.3 N are applied tangentially to a

disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure8.24.

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 :


ICM=1

2MR

2

F1

O

30.0 cm

F2

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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

=12 =RF1RF2=R F1F2 = 0.30 5.6010.3

=

1

2

MR2

=I

1.41=12 5.00 0.30 2

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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 magnitudeP= 3.0 N is applied to the block as

shown in Figure 8.25. Assume the string does not slip on thewheel.

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 :


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a. Free body diagram :

for wheel,

for block,

R=0 .20 m ; I=0.050 kg m2; P=3.0 N; m=2.0 kg

W

TS

NT

Wb

P

a

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b. For wheel,

For block,

By substituting eq. (1) into eq. (2), thus

=I

RT=I T=IR

(1)

F=ma PT=ma (2)

R=0 .20 m ; I=0.050 kg m2; P=3.0 N; m=2.0 kg

a=RPIR =ma andP

I

R =mR

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An object of mass 1.50 kg is suspended

from a rough pulley of radius 20.0 cm by lightstring 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, determine

a. 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.

(Giveng= 9.81 m s-2)

Example 8.18 :


R

1. 50 kg

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a. Free body diagram :

for pulley,

for block,

W

a

T

S =IRT=I = aR

and

RT=I

a

R

T=IaR2 (1)T

m g

F=

mamgT=ma (2)

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a. By substituting eq. (1) into eq. (2), thus

b. By using the relationship between a and , hence

mgIaR2 =ma

R=0 .20 m ; I=0. 020 kg m2; m=1 . 50 kg;

u=0; t=0. 3 s

a=R

1.50 9.81 0.020 a

0.202 =1.50 a

7.36=0.20

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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 m2; m=1 . 50 kg;

u=0; t=0. 3 s

v=uat

v=0 7.36 0 . 3

T=IaR2

T= 0.020 7.36 0.20

2

s=ut12at2

s=01

27.36 0 .3 2

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Exercise 8.4 :


1. Three odd-shaped blocks of chocolate have following massesand 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 perpendicularto the plane

a. through point A, and

b. 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


PHYSICS CHAPTER 8


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96

2 . 00 m s2

T2

T1

Exercise 8.4 :

3. A 5.00 kg object placed on a

frictionless horizontal table isconnected to a string that passesover a pulley and then is fastenedto a hanging 9.00 kg object as inFigure 8.28. The pulley has aradius of 0.250 m and moment of

inertiaI. The block on the table ismoving with a constantacceleration 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. DetermineI.

ANS. : 10.0 N, 70.3 N; 1.88 kg mANS. : 10.0 N, 70.3 N; 1.88 kg m22


PHYSICS CHAPTER 8

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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)

Kr=

1

2 I

2

P=

W=

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8.4 Rotational kinetic energy and power

8.4.1 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

rn

O

v1

v2

v3

vn

Z


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The rigid body has a rotational kinetic energy which is the totaltotal

of kinetic energy of all the particles in the bodyof kinetic energy of all the particles in the body is given by

Kr=12m1 v1212m2 v2

212 m3v 32.. .12 mn vn

2

Kr=1

2m1r1

2

21

2m2r2

2

21

2m3 r3

2

2. ..1

2mn rn

2

2

Kr=122 m1r12m2r22m3r32. . .mnrn2

Kr=1

2I2

Kr=

1

22

i=1

n

miri2 i=1

n

mi ri2=Iand

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From the formula for translational kinetic energy,Ktr

After comparing both equations thus

Forrolling body without slippingrolling body without slipping, the total kinetic energy oftotal kinetic energy of

the body,the body,KKis given by

Ktr=1

2 mv2

is analogous to vvIIis analogous to mm

K=KtrKr

Ktr

: translational kinetic energy

Kr

: rotational kinetic energy

where

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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 sphererolls without slipping from rest to the distance of 75.0 cm and the

inclined surface is smooth, calculate

a. 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 :

ICM=2

5mR

2

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a. From the principle of conservation of energy,

R=0 .15 m ; m=10 .0 kg

Ei=Efmgh=KK=mgs sin25

K=10.0 9.81 0.75 sin25

s=0 .75 m

h=ssin25

v CM 25

R

PHYSICS CHAPTER 8

k


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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=KtrKr K=1

2mv2

1

2 I2 =

v

Rand

K=1

2mv 2

1

2 25 mR2 vR 2

K= 710mv2

31.1=7

1010.0 v2

v=R 2.11=0.15

PHYSICS CHAPTER 8

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The pulley in the Figure 8.30 has a

radius of 0.120 m and a moment ofinertia 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.

(Giveng= 9.81 m s2)

Example 8.20 :

2.00 kg

5.00 kg

7.00 m


PHYSICS CHAPTER 8

5 00 k 2 00 k R 0 120 h 7 00


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105


The moment of inertia of the pulley,

m1=5 . 00 kg ;m

2=2. 00 kg; R=0 .120 m; h=7 . 00 m

I= 0 .055 g 1 cm210

3

kg1 g

104

m2

1 cm2 =5 .5109 kg m2

m2

m1

7.00 m

Initial

m2

m1

7.00 m

v

v

Final

Ei=U1 Ef=Ktr1Ktr2KrU2

PHYSICS CHAPTER 8

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By using the principle of conservation of energy, thus

Ei=EfU

1=K

tr1K

tr2K

rU

2

m1gh=1

2 m1 v2

1

2m2 v2

1

2 I2

m2gh

m1m2 gh=1

2v2 m1m2

1

2I vR

2

5.002.00 9.81 7.00=12v 2 5.002.00 12

5.5109 v

0.120 2

m1=5.00 kg ;m

2=2. 00 kg; R=0 .120 m;

h=7.00 m; I=5.5109

kg m2

PHYSICS CHAPTER 8


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107

Consider a tangential force,Facts 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


F

ds

O

dR

R

dW=Fds

dW=FRd

dW=1

2d W=

1

2d

ds=Rdand

PHYSICS CHAPTER 8

If h i h


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108

If the torque is constant thus

Work-rotational kinetic energy theorem states

W=21

W=1

2d

W=

: torque : change in angular displacement

where

W : work done

W=Kr= KrfKri

W=1

2I2

1

2I0

2

is analogous to the W=Fs

PHYSICS CHAPTER 8


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109

From the definition of instantaneous power,

Caution :

The unitunit ofkinetic energy, work and powerkinetic energy, work and powerin the

rotationalrotational kinematics is samesame as theirunitunit in translationaltranslationalkinematics.

8.4.3 Power,P

P=dWdt

dW=dand

P=d

dtP=

d

dt

=and

is analogous to the P=Fv

PHYSICS CHAPTER 8

E l 8 21


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110

A horizontal merry-go-round has a radius of 2.40 m and a

moment of inertia 2100 kg m2

about a vertical axle through itscentre. A tangential force of magnitude 18.0 N is applied to the

edge of the merry-go- round for 15.0 s. If the merry-go-round is

initially at rest and ignore the frictional torque, determine

a. the rotational kinetic energy of the merry-go-round,

b. the work done by the force on the merry-go-round,c. the average power supplied by the force.

(Giveng= 9.81 m s2)


Example 8.21 :

FR=2 .40 m

PHYSICS CHAPTER 8

S l tiS l ti R 2 40 I 2100 k2

F 18 0 N


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111


a. By applying the relationship between nett torque and angularacceleration, 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 s

1

Kr=1

2I2

Kr=1

22100 0.309

2

R=2 .40 m ; I=2100 kg m2; F=18 .0 N;

t=15.0 s; 0=0

PHYSICS CHAPTER 8

S l tiSolution : R 2 40 I 2100 k2

F 18 0 N


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112


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 t1

2t2

Pav=Wt

W=2.40 18.0 2.32

R=2 .40 m ; I=2100 kg m ; F=18 .0 N;t=15.0 s;

0=0

=01

22.06102 15.0 2

W=RF

Pav=10015.0

PHYSICS CHAPTER 8

Learning Outcome:


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113


Define and useDefine and use angular momentum,angular momentum,

State and useState and use the principle of conservation of angularthe 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


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114

8.5 Conservation of angular momentum

8.5.1 Angular momentum, is defined as the product of the angular velocity of a bodythe 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 vectorquantity.

Its dimension is M LM L22 TT11

The S.I. unit of the angular momentum is kg mkg m22 ss11.

L

where

L=I

L : angular momentumI : moment of inertia of a body : angular velocity of a body

is analogous to the p=mv

PHYSICS CHAPTER 8 The relationship between angular momentum L with linear


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115

The relationship between angular momentum,L with linearmomentum,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 aa vector sum of all the torques acting on arigid body is proportional to the rate of change of angularrigid body is proportional to the rate of change of angularmomentummomentum.

L=

r

p=

r

mv

L=rpsin=mvrsinwhere

:the angle between {rwith {v r : distance from the particle to the rotation axis

F=Fnett=dpdt

=nett=dLdt

PHYSICS CHAPTER 88.5.2 Principle of conservation of angular


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116

states that a total angular momentum of a system about ana total angular momentum of a system about an

rotation axis is constant if no external torque acts on therotation axis is constant if no external torque acts on thesystemsystem.

OR

Therefore

8.5.2 Principle of conservation of angular

momentum

I=constant

=dL

dt=0

dL=0

If the=0

dL=LfLiLi=Lf

and

PHYSICS CHAPTER 8

Example 8 22 :


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117

A 200 kg wooden disc of radius 3.00 m is rotating with angular

speed 4.0 rad s-1

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


1

2

MR2

PHYSICS CHAPTER 8

Solution :Solution : R=3 00 m ; =4 0 rad s1 ; m =200 kg; m =50 kg


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118


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 s ; mw=200 kg; mb=50 kg

Iw=1

2 mwR2

=1

2 200 3.00 2

Iw

0=

I

wI

b

Ib

=mb

R2= 50 3.002

Li=Lf

Iw=900 kg m2

Ib=450 kg m2

900 4.0 =900450

PHYSICS CHAPTER 8

Solution :Solution : R=3 00 m ; =4 0 rad s1 ; m =200 kg; m =50 kg


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119


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 thethe energy is lost in the form of heat from the

friction between the surface of the disc with the bag offriction between the surface of the disc with the bag ofsand.sand.

R=3 .00 m ;0=4 .0 rad s ; mw=200 kg; mb=50 kg

Kri=1

2 Iw02

=1

2 900 4 .0 2

Krf=1

2 IwIb 2

=

1

2 900450 2.67

2

Kri=7200 J

KriKrf

PHYSICS CHAPTER 8

Example 8 23 :


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120

A raw egg and a hard-boiled egg are rotating about the same

axis of rotation with the same initial angular velocity. Explainwhich egg will rotate longer.


The answer is hard-boiled egghard-boiled egg.

Example 8.23 :

PHYSICS CHAPTER 8



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121


ReasonReason

Raw 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 of

From the principle of conservation of angular momentum,

If theIis increases hence its angular velocity, will decreases.

Hard-boiled egg :

The position of the yolk of a hard-boiled 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

Example 8 24 :


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122

A student on a stool rotates freely with an angular speed of 2.95 rev

s

1

. The student holds a 1.25 kg mass in each outstretched arm thatis 0.759 m from the rotation axis. The moment of inertia for the

system of student-stool 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.


Example 8.24 :

0=

2.95 rev

1 s 2 rad

1 rev =18.5 rad s1

=3 .54 rev

1 s 2 rad

1 rev =22. 2 rad s1

PHYSICS CHAPTER 8

Solution :Solution : m=1 25 kg ; 0=18 5 rad s1

; I =5 43 kg m2;


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123Before

0

After

ra

ra


rb

rb

mm

m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ;=22.2 rad s

1;

PHYSICS CHAPTER 8

Solution :Solution : m=1. 25 kg ; 0=18.5 rad s1

; I =5. 43 kg m2;


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124


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

Ii=I

ssI

mIi=Issmrb2mrb2 =I

ss2mr

b2

Ii=5.43 2 1.25 0.759

2=6 . 87 kg m2

If=Iss2mra2

= 5.43 2 1.25 ra

2

Ii

0=I

f

Li=Lf

6.87 18.5 =5.432.5ra222.2

m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ; =22.2 rad s

1;

If=5.432.5r

a2

PHYSICS CHAPTER 8

Solution :Solution : m=1. 25 kg ; 0=18.5 rad s1

; I ss=5. 43 kg m2;


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125


b. The initial rotational kinetic energy is given by

and the final rotational kinetic energy is

Kri=1

2Ii

02

=1

26.87 18.5

2

Kri=1.18103 J

K

rf=

1

2If

2

=1

25.432.5 0.344 222.22

m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ; =22.2 rad s

1;

PHYSICS CHAPTER 8

Exercise 8.5 :


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126


1. A woman of mass 60 kg stands at the rim of a horizontalturntable 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

Exercise 8.5 :


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127

2. Determine the angular momentum of the Earth

a. about its rotation axis (assume the Earth is a uniform solidsphere), and

b. 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 mkg m22 ss11; 2.7 x 10; 2.7 x 104040 kg mkg m22 ss11

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 withangular 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 mkg m22 ss11

1

12ML

2

PHYSICS CHAPTER 8Summary:


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128

Linear Motion Relationship Rotational Motion

m

v=r =ddt

a=dv

dt=

d

dt

F=ma =I

W=Fs W=

v=dsdt

a=r

=rFsin

P=Fv P=

p=mv L=IL=rpsin

II=i=1

n

miri2

PHYSICS CHAPTER 8


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THE END

Next ChapterCHAPTER 9 :

Simple Harmonic Motion (SHM)

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