1. Lecture 17 Goals: Chapter 12 Define center of mass
Analyze rolling motion
Use Work Energy relationships
Introduce torque
Equilibrium of objects in response to forces & torques
Assignment:
HW7 due tomorrow
Wednesday, Exam Review 1

2. Combining translation and rotation
Objects can have translational energy
Objects can have rotational energy
Objects can have both K = ½ m v2 + ½ I w2

3. 1st: A special point for rotation Center of Mass (CM)
A free object will rotate about its center of mass.
Center of mass: Where the system is balanced !
A mobile exploits this centers of mass. m1 m2 + m1 m2 +
mobile

4. System of Particles: Center of Mass
How do we describe the “position” of a system made up of many parts ?
Define the Center of Mass (average position):
For a collection of N individual point like particles whose masses and positions we know:
RCM m2 m1 r2 r1 y x
(In this case, N = 2)

5. Consider the following mass distribution: m at ( 0, 0) 2m at (12,12)
Sample calculation:
Consider the following mass distribution:
m at ( 0, 0)
2m at (12,12)
m at (24, 0)
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
XCM = 12 meters
YCM = 6 meters
(24,0)
(0,0)
(12,12) m 2m
RCM = (12,6)

6. Connection with motion...
An unconstrained rigid object with rotation and translation rotates about its center of mass!
Any point p rotating:
VCM  p

7. Work & Kinetic Energy:
Work Kinetic-Energy Theorem: K = WNET
Applies to both rotational as well as linear motion.
What if there is rolling without slipping ?

8. Same Example : Rolling, without slipping, Motion
A solid disk is about to roll down an inclined plane.
What is its speed at the bottom of the plane ? M q h
v ?

9. Rolling without slipping motion
Again consider a cylinder rolling at a constant speed.
2VCM CM
VCM

10. Motion Again consider a cylinder rolling at a constant speed.
Rotation only
VTang = wR
Both with
|VTang| = |VCM |
Sliding only
2VCM
VCM CM CM CM
VCM
If acceleration acenter of mass = - aR

11. Example : Rolling Motion
A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ?
Use Work-Energy theorem M q h
v ?
Disk has radius R
Mgh = ½ Mv2 + ½ ICM w2 and v =wR
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
v = 2(gh/3)½

12. How do we reconcile force, angular velocity and angular acceleration?

13. From force to spin (i.e., w) ?
A force applied at a distance from the rotation axis gives a torque a
FTangential F q
Fradial r r
FTangential
Fradial
=|FTang| sin q
If a force points at the axis of rotation the wheel won’t turn
Thus, only the tangential component of the force matters
With torque the position & angle of the force matters
NET = |r| |FTang| ≡ |r| |F| sin q

14. Rotational Dynamics: What makes it spin?
FTangential a r
Fradial F
NET = |r| |FTang| ≡ |r| |F| sin q
Torque is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m
NET = r FTang = r m aTang
= r m r a
= (m r2) a
For every little part of the wheel

15. Torque
The further a mass is away from this axis the greater the inertia (resistance) to rotation
FTangential a r
Frandial F
NET = I a
This is the rotational version of FNET = ma
Moment of inertia, I ≡ Si mi ri2 , is the rotational equivalent of mass.
If I is big, more torque is required to achieve a given angular acceleration.

16. NET = I a ≡ |r| |F| sin q Rotational Dynamics F
FTangential F
NET = I a ≡ |r| |F| sin q
Fradial r
A constant torque gives constant angular acceleration if and only if the mass distribution and the axis of rotation remain constant.

17. Torque, like w, has pos./neg. values
Magnitude is given by (1) |r| |F| sin q
(2) |Ftangential | |r|
(3) |F| |rperpendicular to line of action |
Direction is parallel to the axis of rotation with respect to the “right hand rule”
And for a rigid object  = I a
r sin q
line of action
F cos(90°-q) = FTang. r a
90°-q q F F F
Fradial r r r

18. Statics Equilibrium is established when
In 3D this implies SIX expressions (x, y & z) 1

19. Example
Two children (30 kg & 60 kg) sit on a horizontal teeter-totter. The larger child is at the end of the bar and 1.0 m from the pivot point. The smaller child is trying to figure out where to sit so that the teeter-totter remains horizontal and motionless. The teeter-totter is a uniform bar of length 3.0 m and mass 30 kg.
Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)?
The moment of inertia of the bar about the pivot is 30 kg m2.
For the static case:

20. Draw a Free Body diagram (assume g = 10 m/s2)
60 kg
30 kg
0.5 m
300 N
600 N N
Example: Soln.
30 kg
Draw a Free Body diagram (assume g = 10 m/s2)
0 = 300 d x N x 0 – 600 x 1.0
0= 2d + 1 – 4
d = 1.5 m from pivot point

21. Recap
Assignment:
HW7 due tomorrow
Wednesday: review session 1