try to lift the meter stick about an axis at the end by lifting on the far end. keep the “axis”...
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Try to lift the meter stick about an axis at the end by lifting on the far end. KEEP THE “AXIS” ( ) ON THE TABLE!
Is it easy to lift?
F
Now change the position of your force to the middle.
Now change the position to about 20 cm from the axis.
How is the force required changing?
F F
Try to lift the meter stick about an axis at the end by lifting on the far end. KEEP THE “AXIS” ( ) ON THE TABLE!
Is it easy to lift?
F
Now change the position of your force to the middle.
Now change the position to about 20 cm from the axis.
How is the force required changing?
F F
Holt Physics Chapter 8
Rotational Equilibrium and
Dynamics
Apply two equal and opposite forces acting at the center of mass of a stationary meter stick.
Does the meter stick move?
F2
F1
Fext = 0.
F1=F2
Apply two equal and opposite forces acting on a stationary meter stick.
Does the meter stick move?
F
F
The center of mass of the meter stick does not accelerate, so it does not undergo translational motion.
However, the meter stick would begin to rotate about its center of mass.
A torque is produced by a force acting on an extended (not point-like) object.
FThe torque depends on how strong the force is, and where it acts on the object.
Torques cause changes in rotational motion.
Torque is a vector. It is not a forcenot a force,* but is related to force.
*So nevernever set a force equal to a torque!
A
You must always specify your reference axis for calculation of torque. By convention, we indicate that axis with the letter “A” and a dot.
TorqueTorque is a quantity that measures the
ability of a force to rotate an object around some axis.
Torque depends on force and the lever arm.Lever arm (moment arm)is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force.See figure 8-3, page 279
The most torque is produced when
the force is perpendicular to the object.Formula
=Fd(sin)d is the lever arm and is the angle between the lever arm and the force. (If 90º, then sin=1).
See figure 8-5, page 280.
Force is POSITIVE if rotation is counterclockwise.If there is more than one force, add the two resultant torques, using the appropriate signs.EX: Wishbone…sum the two torques
A
F2F1
net = = 1+2 = F1d1 + (-F2d2)
The sign of the net torque will tell you which direction the object will rotate.
A
F2F1
d1 d2
Example 8A, page 281A basketball is being pushed by
two players during tip-off. One player(to the right) exerts a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. The second player(to the left) applies an upward fore of 15 N at a perpendicular distance of 14cm from the axis of rotation. Find the net torque acting on the ball.Direction and sign??? Units???
F1 = -15 N F2 = -11N
d1 = 0.14m d2 = 0.070m
net=?
net= 1+ 2= F1d1+ F2 d 2
(-15N X 0.14m)+(-11N X 0.070m) =
-2.9Nm
Example with an Angle
An upward 34N force is exerted on the right side of a meter stick 0.30 m from the axis of rotation at an angle of 35 degrees. A second downward force of 67N is exerted at an angle of 49 degrees to the meter stick 0.40m to the left of the axis of rotation.
What is the net torque?
34N
67N
35
490.30m
0.40m
Signs???
BOTH are POSITIVE!!
F1 = 34N F2 = 67N
d1 = 0.30m d2 = 0.40m
1= 35 2= 49net=?
net= 1+ 2= F1d1 sin1+ F2d2
sin2
(34NX.30m)sin35+(67NX.40m)sin49=
26Nm
Practice 8A, page 282
EquilibriumComplete equilibrium requires zero
net force and zero net torque.
Translational equilibrium: net force in x and y direction = 0Called 1st condition of equilibrium
∑Fx = 0, ∑Fy = 0Rotational equilibrium: net torque=0Called 2nd condition of equilibrium
∑ = 0
A 45.0m beam that weighs 60.0N is supported in the
center by a cable. The beam is in equilibrium and supports three masses. A
67.0kg mass is on one end, an 89.0kg mass is on the other. A fish is hanging 10.0m from the 67.0 kg mass. What is the mass
(kg!) of the fish and what is the tension (force!) in the
cable.
67kg
89kg60N
45m
10m
?
FT
67kg
89kg60N
Convert to Newtons!
45m
10m
?
FT
657N
873N60N
45m
10m
?
FT
657N
873N60N
Choose AxisChoose center to eliminate a
variable!!
45m
10m
?
FT
657N
873N60N
How far is the fish from the axis?
45m
10m
?
12.5m
FT
657N
873N60N
Calculate Torques and sum to zero
45m
10m
?
12.5m
FT
657N
873N60N
Assign sign to torques
45m
10m
?
12.5m
++++ --00
FT
∑=0=657N(22.5m)+Wf(12.5m)–
873N(22.5m)=0
Wf=4860Nm/(12.5m) = 388.8N
mf=388.8N/9.81m/s2=39.6kg
657N873N60N
45m
10m
39.6kg
12.5m
FT
∑Fy = 0, ∑Fdown = ∑Fup , change all to forces
Fish is 39.6kgX9.81m/s2= 389N
-657N-873N-60N
45m
10m
-389N
12.5m
FT
39.6kg
∑Fy = 0, ∑Fdown = ∑Fup
FT= -∑Fdown =-(-657N- 389N -60N-873N)
=1980N
-657N-873N-60N
45m
10m
-389N
12.5m
FT
F, pillar 2
Fw,
bridge
How to draw a bridge with pillars…
F, pillar 1
Fw, car
Begin Worksheet
Example 8B, page 287-8
A uniform 5.00 m long horizontal beam that weighs 315N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53 degrees with the horizontal, and a 545 N person is standing 1.5 m from the pin. Find the force in the cable, FT, and the force exerted on the beam by the wall, R, if the beam is in equilibrium.
L=5.00 m Fg,beam=315 N =53Fg,person=545 N d=1.50mFT=? R=?Record distances, put weight of object at center of mass and position all forces.
RFT
315 N
1.50 m 53
5.00 m
545 N
The unknowns are R ( Rx, Ry), and FT
Because we have equilibrium, ∑Fx = 0, ∑Fy = 0
Rx - FTcos = 0
Ry + FTsin - Fg,p – Fg,b = 0R
FT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
FTy=FTsin
FTx=FTcos
Rx - FTcos = 0
Ry + FTsin - Fg,p – Fg,b = 0Because there are too many
unknowns, pause (don’t panic) and go to the second condition of equilibrium.
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
Choose an axis and sum the torques…remembering signs for direction of rotation!!
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
Why choose the pin for A?
This eliminates R as a variable!!
0
+
--
=FT L(sin) – Fg,bL/2 –Fg,p d=0Now substitute and solve for FT.
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
FT 5.00m(sin53) – (315N)(5.00m)/2 –(545N)(1.5m)=0
FT= 1606Nm/4.0m
FT= 4.0X102 N
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
We’re not done yet!!
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
Now we can substitute force in the wire (FT) into the Rx and Ry equations to find Rx and Ry, and then solve for R…
RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
Rx - FTcos = 0 Rx = FTcosRx = 400N X cos53 = 240NRy+FTsin-Fg,p–Fg,b=0 Ry= -FTsin53+
Fg,p+Fg,b
Ry = -3.2X102N + 860 N = 540 NR = (Rx
2 + Ry2) = (240N2 + 540N2) =590
N RFT
315 N
1.50 m 53
5.00 m
545 N
Ry
Rx
A
Finish Worksheet!
Homework
Moment of InertiaThe moment of inertia is the resistance of an object to changes in rotational motion about some axis.Similar to mass…mass ( simple inertia) is the measure of resistance to translational motion…
Moment of Inertia depends on the
object’s mass and the distribution of mass around the axis of rotation.The farther the center of mass from the axis of rotation, the more difficult it is to rotate the object, and therefore, the higher the moment of inertia.
Use Table 8-1, page 285
Newton’s 2nd Law
F=ma can be translated to rotational motion.net = I = Iat/r
I = moment of inertia = angular acceleration net = net torqueat=tangential accelerationr=radius
Example page 291
A student tosses dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of inertia of 0.075 kgm2 about the axis, and the length of the forearm is 0.26 m. If the dart has a tangential acceleration of 45m/s2 just before it is released, what is the net torque on the arm and dart?
I = 0.075 kgm2 at=45m/s2
r=0.26m =?=I, where =at/r
=I at/r = 0.075 kgm2 X45m/s2/0.26m=
13Nm
Example 2
A 25 g CD (radius =7.0 cm) is rotating at 100 rev/min. If it stops in 8.5 sec, what is the angular acceleration of the CD? How much torque is required to stop the CD?
r=0.07m, m=0.025kg, t=8.5 sec,i =
100rev/min=100x260=10.5rad/s f= 0 = /t = (0-10.5rad/s)/8.5s
r=0.07m, m=0.025kg, t=8.5 sec,i =
100rev/min=100x260=10.5rad/s f= 0 = /t = (0-10.5rad/s)/8.5s
= -1.2 rad/s2
=I…What is I? Look in table on page 285
Rotating disk is 1/2mr2
= -1.2 rad/s2
=I = 1/2mr2= =1/2(.025kg)(.07)2 -1.2rad/s2
=-7.35X10-5 Nm
8C, page 291
Be ready to use table 8-1 on page 285 to calculate I and use old chapter 7 formulas for quantities like , , s, at and .
Momentum and Rotation
Linear momentum can be translated to angular momentumL = I
L = angular momentumI = moment of inertia – look in the table again (page 285!!)
= angular speed (you may need to use ch. 7 formulas again)
Conservation of Angular Momentum
As in linear momentum, angular momentum is also conserved.
Li = Lf
Example page 293
A 65 kg student is spinning on a merry-go-round that has a mass of 5.25X102 kg and a radius of 2.00 m. She walks from the edge of the merry-go-round toward the center. If the angular speed of the merry-go-round is initially 0.20 rad/sec, what is its angular speed when the student reaches a point 0.50m from the center?
mm = 525 kg ms =65 kg
ri,s= rm =2.00m rf,s = 0.50 m
i =0.20 rad/s f =?
Use conservation of momentumLi = Lf
Lm.i + Ls,i = Lm,f + Ls,f
Need moments of inertia!! Because L = I= Ivt/r
The merry-go-round is a _____The Student is a _____Merry-go-round (I = ½ MR2 )Student (I = MR2 )
Lm.i + Ls,i = Lm,f + Ls,f
½MmRm2i+MsRs,i
2i = ½MmRm
2f+MsRs,f2f
½MmRm2i+MsRs,i
2i = ½MmRm
2f+MsRs,f2f
½525kg(2.00m)2(0.20rad/s)
+65kg(2.00m)2(0.20rad/s) =
½525kg(2.00)2f +65kg(0.50m)2f
Plug it into the calculator and solve for f
f =0.2435rad/s =0.24 rad/s
Keep ch. 7 formulas handy too! Esp. vt=r
Example 2
A comet has a speed of 7.056 X 104 m/s at a distance of 4.95X1010 m. At what distance from the sun would the comet have a speed of 5.0278X104 m/s?
So, I = MR2
Mass is constantω= vt/rRi = 4.95X1010 m Vi=7.056 X 104 m/s Rf= ? Vf=
5.0278X104 m/sMRi
2 vi/ri = MRf2 vf/rf
Ri vi= Rf vf
Rf =Ri vi/ vf
Li=Lf
Point
Mass
Rf = 4.95X1010 m X 7.056 X 104 m/s
5.0278X104 m/sRf= 6.95X1010m
8 D, page 294
Happy Friday!
All labs in?Today we will cover Conservation of Mechanical Energy.
We will skip simple machines and start review problems on Monday for a Wednesday Test.
Kinetic EnergyRotational Kinetic energy (KErot) is the
kinetic energy associated with their angular speed.Formula KErot = ½ I2 = ½ I(vt/r)2 Conservation of Kinetic Energy also
applies…KEtrans + KErot + PEi = KEtrans + KErot + PEf
½ mvi 2 + ½ Ii 2 + mghi = ½ mvf 2 + ½ If 2 + mghf
Be sure to keep track of initial and final conditions as well as angular vs. translational speeds and moments of inertia.
Example page 296A solid ball with a mass of 4.10 kg
and a radius of 0.050m starts from rest at a height of 2.00 m and rolls down a 30 slope. What is the translational speed of the ball when it leaves the incline?
2.00m30
v
hi = 2.00 m m = 4.10 kg
R = 0.050 m vi = 0.0 m/s
= 30.0 hf = 0 m vf = ?
2.00m30
v
I = 2/5MR2What will I be???
hi = 2.00 m m = 4.10 kgR = 0.050 m vi = 0.0 m/s
= 30.0 hf = 0 m vf = ?
½ mvi 2 + ½ Ii 2 + mghi = ½ mvf 2 + ½ If 2 + mghf
2.00m30
v
WE have two variablesSo we need to find oneIn terms of the other…
REMEMBER!!! =vt/r
hi = 2.00 m m = 4.10 kgR = 0.050 m vi = 0.0 m/s
= 30.0 hf = 2.00 m vf = ?
½ mvi 2 + ½ Ii 2 + mghi = ½ mvt,f 2 +½I f 2 +mghf
2.00m30
v
Substitute vt/r intoThe equation for f
(vt,f /r)
4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf 2 +½I (vf/0.050m) 2
I = 2/5MR2
4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf 2 +½(2/5MR2) (vf/0.050m) 2
PLUG IN THE NUMBERS
80.442kgm2/s2 = 2.05kg vf 2 + 0.82kg vf 2
2.87kg vf 2 = 80.442kgm2/s2
vf 2 =28 m2/s2
vf = 5.29 m/s
4.10kg(9.81m/s 2) (2.00m) =½ 4.10kg vf 2 +½(2/5)(4.10kg)(0.050m) 2 (vf/0.050m) 2
Example 2
A 3.5 kg spherical potato (radius .070m) is kicked up a 30 degree slope at a speed of 5.4 m/s. What distance along the slope did the potato roll before it stopped?
mp= 3.5kg, vi,p= 5.4m/s θ=30°hf=? Hypotenuse (slope dist.)= ? ½ mvi 2 + ½ Ii 2 +mghi = ½ mvf 2 + ½ If 2 + mghf
*Substitute vt/r for ω
½ mvi 2 + ½ Ii 2 = mghf
*Solve for hf
hf=½ mvi 2 + ½ I(vt /r)2 =(51.03 + 20.412)/34.335
mg
hf=2.08m
D=2.08/sin30=4.2m
(vt,i/r)2
(vt,i/r)2
distance heig
ht30
Kinetic Energy
Practice 8E, page 297