1. General Physics I Rotational Motion

2. Rotation
Up until now we have gracefully avoided dealing with the rotation of objects.
We have studied objects that slide, not roll.
We have assumed pulleys are without mass.
Rotation is extremely important, however, and we need to understand it!
Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics.

3. System of Particles
Until now, we have considered the behavior of very simple systems (one or two masses).
But real life is usually much more interesting!
For example, consider a simple rotating disk.
An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object!

4. System of Particles: Center of Mass
How do we describe the “position” of a system made up of many parts?
We define the Center of Mass
The center of mass is where the system is balanced!
Building a mobile is an exercise in finding centers of mass. m1 m2 + m1 m2 +

5. System of Particles: Center of Mass
We can use intuition to find the location of the center of mass for symmetric objects that have uniform density:
It will simply be at the geometrical center ! + CM + + + + +

6. Rotation & Kinetic Energy
Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).
The kinetic energy of this system will be the sum of the kinetic energy of each piece: r1 r2 r3 r4 m4 m1 m2 m3 

7. Rotation & Kinetic Energy...
So: but vi = ri r1 r2 r3 r4 m4 m1 m2 m3  v4 v1 v3 v2
Which we write as:
...the moment of inertia
about the rotation axis
I has units of kg m2.

8. Moment of Inertia
So where
Notice that the moment of inertia I depends on the distribution of mass in the system.
The further the mass is from the rotation axis, the bigger the moment of inertia.
For a given object, the moment of inertia will depend on where we choose the rotation axis.
We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics!

9. Some Sample Moments of Inertia
Thin hoop (or cylinder) of mass M and
radius R, about an axis through its center,
perpendicular to the plane of the hoop. R
Thin rod of mass M and length L, about
a perpendicular axis through its center. L
Thin rod of mass M and length L, about
a perpendicular axis through its end. L

10. Some More Samples... Solid sphere of mass M and radius R,
about an axis through its center. R R
Solid disk or cylinder of mass M and
radius R, about a perpendicular axis
through its center; a pulley!

11. Rolling... v 2v v Linear Motion Only Linear Motion + Rotation  v
Rotation Only
Where v = R 

12. Connection with Linear motion...
So for a solid object which rotates about its center and which is also moving with linear velocity: 
VCM

13. Example of Rolling Motion
Objects of different I rolling down an inclined plane:
K = - U = Mgh
v = 0
= 0
K = 0 R M h
v = R

14. Consider the conservation of energy of the system!!
Rotations
Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other.
If both are placed at the top of the same ramp and released (they roll without slipping), which is moving faster (linear velocity) at the bottom?
(a) bigger one
(b) smaller one
(c) same
Consider the conservation of energy of the system!!

15. Rotations - Solution
OK…let’s starts with our conservation of energy equation and go from there….

16. Torque
If an applied force causes a mass to rotate (circular motion) then we have defined a torque:
 = rF
= r F sin 
= r sin  F F F  Fr r

17. Rotational Dynamics: What makes it spin?
Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant:
a = r
Now use Newton’s 2nd Law in the  direction:
F = ma = mr
rF = mr2 r a  F m  F
Multiply by r :

18. Rotational Dynamics: What makes it spin?
So…. rF = mr2 &  = rF &
Thus:
Torque has a direction:
+ if it tries to make the system spin CCW.
- if it tries to make the system spin CW. r a  F m  F

19. Rotational Dynamics: What makes it spin?
 NET = I
This is the rotational analogue of FNET = ma
Torque is the rotational analogue of force:
The amount of “twist” provided by a force.
Moment of inertia I is the rotational analogue of mass.
If I is big, more torque is required to achieve a given angular acceleration.
Torque has units of kg m2/s2 = (kg m/s2) m = Nm.

20. Atwoods Machine with Massive Pulley
A pair of masses are hung over a massive disk-shaped pulley as shown.
What are the equations necessary to solve for the unknowns in the problem?
Three objects have mass so we will have three equations.
What does that imply about the number of unknowns?
First step is to complete the free-body diagram! M R m1 m2

21. Atwood’s Machine with Massive Pulley...
For the hanging masses use F = ma M
For the pulley use  = I R m2 m2 m1 m1

22. What About Statics? In general, we can use the two equations
to solve any statics problem.
When choosing axes about which to calculate torque, we can be clever and make the problem easy....

23. Using Torque in Static Cases:
Now consider a plank of mass M suspended by two strings as shown. We want to find the tension in each string:
L/2
L/4 M
x cm T1 T2 Mg y x

24. Statics
A 1 kg ball is hung at the end of a rod 1 m long. The system balances at a point on the rod 0.25 m from the end holding the mass.
What is the mass of the rod?
(a) 0.5 kg (b) 1 kg (c) 2 kg
1 m
1 kg

25. Statics
A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1kg.
What is the total mass of the mobile?
(a) 5 kg (b) 6 kg (c) 7 kg
1 m
2 m
1 kg
1 m
3 m

26. Angular momentum of a rigid body about a fixed axis:
We define the rotational analogue of momentum p to be angular momentum
Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle:
Using vi =  ri , we get: v1 m2 y r2 r1 m1 I  x v2
Analogue of p = mv!!  r3 m3 v3

27. Example: Rotating Table...f Ii If Li Lf

28. End of Rotational Motion Lecture