1. Torque and Angular Momentum
Chapter 8
Torque and Angular Momentum

2. Torque and Angular Momentum
Rotational Kinetic Energy
Rotational Inertia
Torque
Work Done by a Torque
Equilibrium (revisited)
Rotational Form of Newton’s 2nd Law
Rolling Objects
Angular Momentum
MFMcGraw
Chap08-Torque-Revised 3/6/2010

3. Rotational KE and Inertia
For a rotating solid body:
For a rotating body vi = ri where ri is the distance from the rotation axis to the mass mi.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 1, 11, and 13.
MFMcGraw
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4. Chap08-Torque-Revised 3/6/2010
Moment of Inertia
The quantity
is called rotational inertia or moment of inertia.
Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in the table in the textbook for extended bodies.
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5. Chap08-Torque-Revised 3/6/2010
Moments of Inertia
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6. Chap08-Torque-Revised 3/6/2010
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Moment of Inertia
Example: The masses are m1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless.
Q: (a) Find the moment of inertia of the system below.  r1 r2 m1 m2
r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively.
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8. Chap08-Torque-Revised 3/6/2010
Moment of Inertia
Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m.
(b) What is the moment of inertia if the axis is moved so that is passes through m1?
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9. Chap08-Torque-Revised 3/6/2010
Moment of Inertia
What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it?
From the table:
MFMcGraw
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10. Chap08-Torque-Revised 3/6/2010
A torque is caused by the application of a force, on an object, at a point other than its center of mass or its pivot point.
hinge
Push
Q: Where on a door do you normally push to open it?
A: Away from the hinge.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Question 16.
A rotating (spinning) body will continue to rotate unless it is acted upon by a torque.
MFMcGraw
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11. Chap08-Torque-Revised 3/6/2010
Torque method 1:
Hinge end F 
Top view of door
r = the distance from the rotation axis (hinge) to the point where the force F is applied.
F is the component of the force F that is perpendicular to the door (here it is Fsin).
MFMcGraw
Chap08-Torque-Revised 3/6/2010

12. Torque The units of torque are Newton-meters (Nm) (not joules!).
By convention:
When the applied force causes the object to rotate counterclockwise (CCW) then  is positive.
When the applied force causes the object to rotate clockwise (CW) then  is negative.
CW is clockwise and CCW is counterclockwise.
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13. Chap08-Torque-Revised 3/6/2010
Torque method 2:
r is called the lever arm and F is the magnitude of the applied force.
Lever arm is the perpendicular distance to the line of action of the force.
MFMcGraw
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14. Chap08-Torque-Revised 3/6/2010
Line of action of the force 
Hinge end F  r
Top view of door
Lever arm
The torque is:
Same as before
MFMcGraw
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15. Chap08-Torque-Revised 3/6/2010
Torque Problem
The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine.
What magnitude torque does the cord apply to the drum?
F=75 N
R=6.00 cm
MFMcGraw
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16. Chap08-Torque-Revised 3/6/2010
Torque Problem
Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the bar.
45
10
30
F1=25 N
F3=20 N
F2=30 N X
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17. Chap08-Torque-Revised 3/6/2010
Torque Problem
Lever arm for F3
Lever arm for F2
45
10
30
F1=25 N
F3=20 N
F2=30 N X
This is a fairly complex figure. It will probably be useful if you walk your students through its construction.
The lever arms are:
MFMcGraw
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18. Chap08-Torque-Revised 3/6/2010
Torque Problem
The torques are:
The net torque is Nm and is the sum of the above results.
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19. Chap08-Torque-Revised 3/6/2010
Work done by the Torque
The work done by a torque  is
where  is the angle (in radians) that the object turns through.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 17 and 20.
Following the analogy between linear and rotational motion:
Linear Work is Force x displacement. In the rotational picture force becomes torque and displacement becomes the angle
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20. Chap08-Torque-Revised 3/6/2010
Work done by the Torque
A flywheel of mass 182 kg has a radius of 0.62 m (assume the flywheel is a hoop).
(a) What is the torque required to bring the flywheel from rest to a speed of 120 rpm in an interval of 30 sec?
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Work done by the Torque
(b) How much work is done in this 30 sec period?
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22. Chap08-Torque-Revised 3/6/2010
Equilibrium
The conditions for equilibrium are:
Linear motion
Rotational motion
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 8, 9, 12, and 14.
For motion in a plane we now have three equations to satisfy.
MFMcGraw
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23. Chap08-Torque-Revised 3/6/2010
Using Torque
A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at a 35 angle with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of the sign is N.
What is the tension in the cable and what are the horizontal and vertical forces exerted on the boom by the hinge?
MFMcGraw
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24. Chap08-Torque-Revised 3/6/2010
Using Torque
This is important! You need two components for F, not just the expected perpendicualr normal force.
FBD for the bar: T
wbar Fy Fx  X
Fsb x y
Apply the conditions for equilibrium to the bar:
Fsb is the force that the sign exerts on the boom. Drawing a free body diagram for the sign and applying Newton’s second and third laws gives the magnitude of this force as the weight of the sign. L is the length of the boom (given as 3.00 m) and x is the distance from the hinge to the location of the cable (given as 2.38 m). Note: the lines of action and the lever arms are omitted from the figure for clarity.
MFMcGraw
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25. Chap08-Torque-Revised 3/6/2010
Using Torque
Equation (3) can be solved for T:
Equation (1) can be solved for Fx:
Equation (2) can be solved for Fy:
MFMcGraw
Chap08-Torque-Revised 3/6/2010

26. Equilibrium in the Human Body
Find the force exerted by the biceps muscle in holding a one liter milk carton with the forearm parallel to the floor. Assume that the hand is 35.0 cm from the elbow and that the upper arm is 30.0 cm long.
The elbow is bent at a right angle and one tendon of the biceps is attached at a position 5.00 cm from the elbow and the other is attached 30.0 cm from the elbow.
The weight of the forearm and empty hand is 18.0 N and the center of gravity is at a distance of 16.5 cm from the elbow.
MFMcGraw
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27. Chap08-Torque-Revised 3/6/2010
MCAT type problem Fb w
Fca
“hinge” (elbow joint) x y
Fca is the force that the carton exerts on the arm. Drawing a free body diagram for the carton and applying Newton’s second and third laws gives the magnitude of this force as the weight of the carton. x1 is the lever arm for the “biceps force” (given as 5.00 cm; assumes the force Fb is nearly perpendicular to the forearm since no angle is given), x2 is the lever arm for the “weight of the forearm” (given as 16.5 cm), and x3 is the lever arm for the “force of the carton on the arm” (given as 35.0 cm). Note: the lines of action and the lever arms are omitted for clarity.
MFMcGraw
Chap08-Torque-Revised 3/6/2010

28. Newton’s 2nd Law in Rotational Form
Compare to
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Question 18.
MFMcGraw
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29. Chap08-Torque-Revised 3/6/2010
Rolling Object
A bicycle wheel (a hoop) of radius 0.3 m and mass 2 kg is rotating at 4.00 rev/sec. After 50 sec the wheel comes to a stop because of friction.
What is the magnitude of the average torque due to frictional forces?
MFMcGraw
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30. Chap08-Torque-Revised 3/6/2010
Rolling Objects
An object that is rolling combines translational motion (its center of mass moves) and rotational motion (points in the body rotate around the center of mass).
For a rolling object:
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 2, 3, 4, and 5.
If the object rolls without slipping then vcm = R.
Note the similarity in the form of the two kinetic energies.
MFMcGraw
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31. Chap08-Torque-Revised 3/6/2010
Rolling Example
Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height.
Which object reaches the bottom of the ramp first? h 
This we know - The object with the largest linear velocity (v) at the bottom of the ramp will win the race.
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32. Chap08-Torque-Revised 3/6/2010
Rolling Example
Apply conservation of mechanical energy:
Solving for v:
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33. Chap08-Torque-Revised 3/6/2010
Rolling Example
Example continued:
Note that the mass and radius are the same.
The moments of inertia are:
For the disk:
Since Vsphere> Vdisk the sphere wins the race.
For the sphere:
Compare these to a box sliding down the ramp.
MFMcGraw
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34. Chap08-Torque-Revised 3/6/2010
The Disk or the Ring?
MFMcGraw
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35. How do objects in the previous example roll?y
FBD:
Both the normal force and the weight act through the center of mass so
 = 0.
This means that the object cannot rotate when only these two forces are applied. x
The round object won’t rotate, but most students have difficulty imagining a sphere that doesn’t rotate when moving down hill.
MFMcGraw
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36. Chap08-Torque-Revised 3/6/2010
Add Friction y x N w Fs 
FBD:
Also need acm = R and
The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.13.)
It is the addition of static friction that makes an object roll.
MFMcGraw
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37. Chap08-Torque-Revised 3/6/2010
Angular Momentum
Units of p are kg m/s
Units of L are kg m2/s
When no net external forces act, the momentum of a system remains constant (pi = pf)
When no net external torques act, the angular momentum of a system remains constant (Li = Lf).
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 6, 7, and 19.
MFMcGraw
Chap08-Torque-Revised 3/6/2010

38. Chap08-Torque-Revised 3/6/2010
Angular Momentum
Units of p are kg m/s
Units of L are kg m2/s
When no net external forces act, the momentum of a system remains constant (pi = pf)
When no net external torques act, the angular momentum of a system remains constant (Li = Lf).
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 6, 7, and 19.
MFMcGraw
Chap08-Torque-Revised 3/6/2010

39. Angular Momentum Example
A turntable of mass 5.00 kg has a radius of m and spins with a frequency of rev/sec. Assume a uniform disk.
What is the angular momentum?
MFMcGraw
Chap08-Torque-Revised 3/6/2010

40. Angular Momentum Example
A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended.
What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2?
The skater is on ice, so we can ignore external torques.
MFMcGraw
Chap08-Torque-Revised 3/6/2010

41. The Vector Nature of Angular Momentum
Angular momentum is a vector. Its direction is defined with a right-hand rule.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 8, Questions 10 and 15.
MFMcGraw
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42. Chap08-Torque-Revised 3/6/2010
The Right-Hand Rule
Curl the fingers of your right hand so that they curl in the direction a point on the object moves, and your thumb will point in the direction of the angular momentum.
Angular Momentum is also an example of a vector cross product
MFMcGraw
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43. The Vector Cross Product
The magnitude of C
C = ABsin(Φ)
The direction of C is perpendicular to the plane of A and B.
Physically it means the product of A and the portion of B that is perpendicular to A.
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44. The Cross Product by Components
Since A and B are in the x-y plane A x B is along the z-axis.
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45. Memorizing the Cross Product
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46. Chap08-Torque-Revised 3/6/2010
The Gyroscope Demo
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47. Chap08-Torque-Revised 3/6/2010
Angular Momentum Demo
Consider a person holding a spinning wheel. When viewed from the front, the wheel spins CCW.
Holding the wheel horizontal, they step on to a platform that is free to rotate about a vertical axis.
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48. Chap08-Torque-Revised 3/6/2010
Angular Momentum Demo
Initially, nothing happens.
They then move the wheel so that it is over their head. As a result, the platform turns CW (when viewed from above).
This is a result of conserving angular momentum.
MFMcGraw
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49. Chap08-Torque-Revised 3/6/2010
Angular Momentum Demo
Initially there is no angular momentum about the vertical axis. When the wheel is moved so that it has angular momentum about this axis, the platform must spin in the opposite direction so that the net angular momentum stays zero.
Is angular momentum conserved about the direction of the wheel’s initial, horizontal axis?
MFMcGraw
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50. Chap08-Torque-Revised 3/6/2010
It is not.
The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here.
MFMcGraw
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51. Chap08-Torque-Revised 3/6/2010
Summary
Rotational Kinetic Energy
Moment of Inertia
Torque (two methods)
Conditions for Equilibrium
Newton’s 2nd Law in Rotational Form
Angular Momentum
Conservation of Angular Momentum
MFMcGraw
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52. Chap08-Torque-Revised 3/6/2010X Z Y
MFMcGraw
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