1. Foundations of Physics
CPO Science
Foundations of Physics
Chapter 9
Unit 3, Chapter 9

2. Unit 3: Motion and Forces in 2 and 3 Dimensions
Chapter 9 Torque and Rotation
9.1 Torque
9.2 Center of Mass
9.3 Rotational Inertia

3. Chapter 9 Objectives Calculate the torque created by a force.
Solve problems by balancing two torques in rotational equilibrium.
Define the center of mass of an object.
Describe a technique for finding the center of mass of an irregularly shaped object.
Calculate the moment of inertia for a mass rotating on the end of a rod.
Describe the relationship between torque, angular acceleration, and rotational inertia.

4. Chapter 9 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

5. 9.1 Torque
Key Question:
How does force create rotation?

6. 9.1 Torque A torque is an action that causes objects to rotate.
Torque is not the same thing as force.
For rotational motion, the torque is what is most directly related to the motion, not the force.

7. 9.1 Torque
Motion in which an entire object moves is called translation.
Motion in which an object spins is called rotation.
The point or line about which an object turns is its center of rotation.
An object can rotate and translate.

8. 9.1 Torque
Torque is created when the line of action of a force does not pass through the center of rotation.
The line of action is an imaginary line that follows the direction of a force and passes though its point of application.

9. 9.1 Torque
To get the maximum torque, the force should be applied in a direction that creates the greatest lever arm.
The lever arm is the perpendicular distance between the line of action of the force and the center of rotation

11. 9.1 Torque
Lever arm length (m)
τ = r x F
Torque (N.m)
Force (N)

12. 9.1 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 so the lever arm is 30 cm.

13. 9.1 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?

14. 9.1 Rotational Equilibrium

15. 9.1 Calculate using equilibrium
A boy and his cat sit on a seesaw.
The cat has a mass of 4 kg and sits 2 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?
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.

16. 9.1 Calculate using equilibrium
A boy and his cat sit on a seesaw.
The cat has a mass of 4 kg and sits 2 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?
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.

17. *Students read Section 9.2 AFTER Investigation 9.2
9.2 Center of Mass
Key Question:
How do objects balance?
*Students read Section 9.2 AFTER Investigation 9.2

18. 9.2 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.

20. 9.2 Finding the center of gravity
The center of gravity of an irregularly shaped object can be found by suspending it from two or more points.
For very tall objects, such as skyscrapers, the acceleration due to gravity may be slightly different at points throughout the object.

21. 9.2 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 the entire region surrounded by the actual supports.
An object will topple over if its center of mass is not above its area of support.

23. *Students read Section 9.3 AFTER Investigation 9.3
9.3 Rotational Inertia
Key Question:
Does mass resist rotation the way it resists acceleration?
*Students read Section 9.3 AFTER Investigation 9.3

24. 9.3 Rotational Inertia
Inertia is the name for an object’s resistance to a change in its motion (or lack of motion).
Rotational inertia is the term used to describe an object’s resistance to a change in its rotational motion.
An object’s rotational inertia depends not only on the total mass, but also on the way mass is distributed.

25. 9.3 Rotational Inertia
A rotating mass on a rod can be described with variables from linear or rotational motion.

26. 9.3 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.

27. 9.3 Moment of Inertia
The sum of mr2 for all the particles of mass in a solid is called the moment of inertia (I).
A solid object contains mass distributed at different distances from the center of rotation.
Because rotational inertia depends on the square of the radius, the distribution of mass makes a big difference for solid objects.

28. 9.3 Moment of Inertia
The moment of inertia of some simple shapes rotated around axes that pass through their centers.

29. 9.3 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.

30. Application: Bicycle Physics