1. Chapter 8 Torque and Rotation
DE Physics
Chapter 8 Torque and Rotation
8.2 Torque and Stability
6.5 Center of Mass
8.3 Rotational Inertia Dorsey, Adapted from CPO Science

2. Chapter 8 Objectives Calculate the torque created by a force.
Solve problems by balancing two torques in rotational equilibrium.
Using center of mass to find the torque
Calculate the moment of inertia for rotation mass.
Know and use the Moment of Inertia of common objects
Describe the relationship between torque, angular acceleration, and rotational inertia.

3. Chapter 8 Vocabulary Terms
torque
center of mass
angular acceleration
rotational inertia
rotation
translation
center of rotation
rotational equilibrium
lever arm
center of gravity
moment of inertia
line of action

4. 8.2 Torque
Key Question:
How does force create rotation?

5. 8.2 Torque A torque is an action that causes objects to rotate.
Torque is not the same thing as force.

6. Torque Motion in which an entire object moves is called translation.
Motion in which an object spins is called rotation.

7. Torque
The point or line about which an object turns is its center of rotation.
An object can rotate and translate.

8. Torque
The line of action: Goes through the point of application in the direction of the applied force
Torque is created when the line of action….???

9. Let’s see. Put a book on the table. Push it in the middle
Then push it on the edge
How does it move both times?

10. Torque
Torque is created when the line of action, does not pass through the center of rotation.

11. Another Example
When you came in the classroom you used torque to get in.
What did you have to do to get in the room?

12. Opening a Door Where is the handle with respect to the hinges?
Go to the door as a class:
try opening it really close to the hinges
then as far as you can go.

13. Opening a Door Which was easier
Opening from the edge or near the hinges?

14. Torque
Maximum torque happens when pulling perpendicular to the shaft

16. t = r x F Torque Lever arm length (m) Torque (N.m) Force (N)
Right hand rule

17. Thumb pointing out of page is + Thumb into page is -
The right hand rule:
Curl your right hand following the torque arrow with your thumb in the middle.
Thumb pointing out of page is + Thumb into page is -
Right hand rule

18. Calculate a torque
A force of 50. newtons is applied to a wrench that is 30. centimeters long.
1) You are asked to find the torque.
2) You are given the force and lever arm.
3) The formula that applies is τ = rF.
4) Solve: τ = (-50 N)(0.3 m) = -15 N.m
Calculate the torque if the force is applied perpendicular to the wrench, making 30 cm lever arm.

19. The torque is negative following the right hand rule.
Calculate a torque
A force of 50. newtons is applied down, counter clockwise, to a wrench in a way with a 30. cm long lever arm.
τ = (-50. N)(0.30 m) = -15 N.m
The torque is negative following the right hand rule.
1) You are asked to find the torque.
2) You are given the force and lever arm.
3) The formula that applies is τ = rF.
4) Solve: τ = (-50 N)(0.3 m) = -15 N.m

20. When the force and lever arm are NOT perpendicular

21. 9.1 Calculate a torque
Perpendicular Force 50. N turns this bolt. What force is needed at 30. degrees:
Knowns: r = .20 m Angle2 = 30 deg
F1 = 50. N F2 = ?
Angle1= 0 deg
Eq: τ = rF.
1) You are asked to find the force.
2) You are given the force and lever arm for one condition.
3) The formula that applies is τ = rF.
4) Solve: The torque required to loosen the bolt τ = (50 N)(0.2 m) = 10 N.m
To get the same torque with a force applied at 30 degrees:
10 N.m = F × (0.2 m)cos30o = F
F = 10 N.m ÷ = 58 N. It takes a larger force.

22. Calculate a torque
Perpendicular Force 50 N turns this bolt. What force is needed at 30 degrees:
Knowns: Eq: τ = rF.
r = .20 m Angle = 30 deg
F = 50. N F(perpen) = ?
Torque required to break τ = (50 N)(0.2 m) = 10 N.m
To get the same torque with a F at 30 degrees:
10 N.m = F × (0.2 m)cos30o
10 N.m = F F = 58 N : More force is required
1) You are asked to find the force.
2) You are given the force and lever arm for one condition.
3) The formula that applies is τ = rF.
4) Solve: The torque required to loosen the bolt τ = (50 N)(0.2 m) = 10 N.m
To get the same torque with a force applied at 30 degrees:
10 N.m = F × (0.2 m)cos30o = F
F = 10 N.m ÷ = 58 N. It takes a larger force.

23. Rotational Equilibrium
When an object is in rotational equilibrium, the net torque applied to it is zero.
Rotational equilibrium is often used to determine unknown forces.

24. Calculate using equilibrium
A boy and his cat sit on a seesaw.
The cat has a mass of 4.0 kg and sits 2.0 m from the center of rotation.
If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

25. Calculate using equilibrium
The cat, mass of 4 kg sits 2 m from the center of rotation. Where should a boy of mass 50 kg sit to balance the system
Solve: τcat = (2 m)(4 kg)(9.8 N/kg) = N-m
τboy = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque = zero.
d = 0
d = 0.16 m The boy must sit 16 cm from the center.
1) You are asked to find the boy’s lever arm.
2) You are given the two masses and the cat’s lever arm.
3) Torque, τ = rF weight: F = mg
In equilibrium the net torque must be zero.
4) Solve: For the cat,
τ = (2 m)(4 kg)(9.8 N/kg) = N-m
For the boy:
τ = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque must be zero.
d = 0
d = 0.16 m
The boy must sit quite close (16 cm) to the center.

27. 6.5 Center of Mass
Key Question:
How do objects balance?

28. Center of Mass
There are three different axes about which an object will naturally spin.
The point at which the three axes intersect is called the center of mass.

29. Finding the center of mass
Center of mass for all objects can be found by spinning the object and finding the intersection of the three spin axes.
There is not always material at an object’s center of mass.

31. Finding the center of gravity
Center of gravity is different from center of mass.
For very tall objects, such as skyscrapers, the acceleration due to gravity may be slightly different at points throughout the object.

32. Balance and center of mass
For an object to remain upright, its center of gravity must be above its area of support.
The area of support includes all the area within the supports.
An object will topple over if its center of mass is not above its area of support.

33. Use the center of mass toppler

34. Center of mass and people:
Start at 1:00

35. Rotational Inertia Key Question:
Does mass resist rotation the same way it resists linear acceleration?

36. Rotational Inertia Inertia is resistance to a change in its motion
Rotational inertia describes an object’s resistance to a change in its rotational motion.
An object’s rotational inertia depends on the total mass and the way mass is distributed.
Look at the pictures on the right. Mass further means more inertia

37. Watch this!
Mass on the end means slower

38. Watch this!

39. Linear and Angular Acceleration
Angular acceleration (kg)
Linear
acceleration
(m/sec2)
a = a r
Radius of motion
(m)

40. Rotational Inertia
Rotational motion’s equation has force replaced by the torque about the center of rotation.
The linear acceleration is replaced by the angular acceleration.

41. Rotational Inertia
A rotating mass on a rod can be described with variables from linear or rotational motion.

42. Rotational Inertia
The product of mass × radius squared (mr2) is the rotational inertia for a point mass where r is measured from the axis of rotation.

44. Moment of Inertia Is analogous to the rotational mass
Greater the moment of inertia: the harder it is to change its motion

45. Moment of Inertia For just one particle it is I = mr2
Each solid object has mass distributed at different distances from the center of rotation.
Mass distribution makes a big difference for rotational inertia because of the r2 relationship

46. Moment of Inertia
The moment of inertia of some simple shapes rotated around their center axes
You need to write these six down and know them!!!
Y is all that matters

47. Watch This: Why is one faster than the other.

48. Rotation and Newton's 2nd Law
If you apply a torque to a wheel, it will spin in the direction of the torque.
The greater the torque, the greater the angular acceleration.

49. Solving Inertia Example
Two point mass of 1 kg are on opposite ends of a 2 m long massless rod.
If spun around the center, what is the rotational inertia, I.
I=m1r2 + m2r2 = 1*12+1*12 = 2 kg*m2

50. Solving Inertia Example
Two point mass of 1 kg are on opposite ends of a 2 m long massless rod.
If spun at one end, what is I.
I=m1r2 = 1*22= 4 kg*m2

51. 12.3 Angular Momentum
Momentum resulting from an object moving in linear motion is linear momentum.
Momentum resulting from the rotation (or spin) is called angular momentum.

52. 12.3 Conservation of Angular Momentum
Angular momentum is conserved!
The total angular momentum of a closed system stays the same.

53. 12.3 Calculating angular momentum
Moment of inertia
(kg m2)
L = I w
Angular
momentum
(kg m/sec2)
Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity.
Angular
velocity
(rad/sec)

55. Calculating angular momentum
A 1.0 m, 1.0 kg straight bar and a hoop with a radius of 0.16 m are spun around the center with an angular velocity of 1 rad/sec.
Calculate the angular momentum of each and decide which would be harder to stop.
You are asked for angular momentum.
You are given mass, shape, and angular velocity.
Hint: both rotate about y axis.
Use L= I, Ihoop = mr2, Ibar = 1/12 ml2

56. Calculating angular momentum
Solve hoop: Ihoop= (1 kg) (0.16 m)2 = kg m2
Lhoop= (1 rad/s) (0.026 kg m2) = kg m2/s
Solve bar: Ibar = (1/12)(1 kg) (1 m)2 = kg m2
Lbar = (1 rad/s) (0.083 kg m2) = kg m2/s
The bar has more than 3x the angular momentum of the hoop, so it is harder to stop.

57. Stability

58. Slides after this are extra untested concepts

59. 12.3 Gyroscopes angular momentum
A gyroscope is a device that contains a spinning object with a lot of angular momentum.
Gyroscopes can do amazing tricks because they conserve angular momentum.
For example, a spinning gyroscope can easily balance on a pencil point.

60. 12.3 Gyroscopes angular momentum
A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the spacecraft in three dimensions.
An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in space.

61. Jet Engines
Nearly all modern airplanes use jet propulsion to fly. Jet engines and rockets work because of conservation of linear momentum.
A rocket engine uses the same principles as a jet, except that in space, there is no oxygen.
Most rockets have to carry so much oxygen and fuel that the payload of people or satellites is usually less than 5 percent of the total mass of the rocket at launch.

62. Force on bridge supports
ewdyn/html/static_forces_bridge.htm
Play with this for no longer than 3 min.
Notice how the forces on the pillars change as the truck moves across the bridge.

63. Force on bridge supports
Find the fraction of where the center of mass is to the total length of the bridge
That is the fraction of the weight that is on the further pillar
The closer pillar takes the rest of the weight

64. Force on bridge supports
You can also sum the torques of what is on the bridge and the torque the CM of the bridge applies.

65. Rotational Equilibrium
When an object is in rotational equilibrium, the net torque applied to it is zero.
Rotational equilibrium is often used to determine unknown forces.
What are the forces (FA, FB) holding the bridge up at either end?

66. Rotational Equilibrium