1. Chapter 11
Rotational Dynamics

2. Important quantities

i
i
i R
m
2

i
i
iv
m


i
i
i
i
i
m
R
m
Moment of inertia, I. Used in t=Ia. So, I
divided by time-squared has units of force
times distance (torque).
Total linear momentum is the sum
of the individual momenta.
Center-of-mass is a distance. Has
to have units of meters.

3. Rotational Relationships
2
2
1

a
t


a



I
K
I
I
L
t
t


















Position
Velocity
Acceleration
Momentum
Force/Torque
Kinetic Energy
t
t
t
a


a








0
2
0
2
1
Master equations:
One-to-one
correspondence of
rotational equations to
linear equations.

4. Torque
 
 
 
a
a
a
t
I
mr
r
r
m
r
ma
Fr





2

5. Force and Torque Combined: What is the acceleration of a
pulley with a non-zero moment of inertia?
R
a
I
I
TR
R
a
TR
I





a
a
t
a
t
Torque relation for pulley:
ma
T
mg 

Force Relation for Mass
Put it together:











2
2
1
mR
I
ma
mg
R
a
I
T
ma
T
mg














2
1
1
mR
I
g
a
NOTE: Positive down!

6. How high does it go?
Use energy conservation.
Initial Kinetic Energy:
Initial Potential Energy:
Final Kinetic Energy:
Final Potential Energy:
Zero
Zero
2
2
2
1
2
1

I
mv
KI 

mgH
UF 
Use v=R, and then set Initial Energy = Final Energy.










2
2
2
2
2
1
2
2
1
2
1
mR
I
g
v
H
mgH
R
v
I
mv

7. Angular Momentum
Angular momentum is “conserved” (unchanged), in the
absence of an applied torque.
t
L





t



I
L 

8. Comparison of Linear and Angular Momentum
m
P
K
t
P
F
V
m
P
2
2









I
L
K
t
L
I
L
2
2









t

Linear momentum is conserved in absence of an applied force.
Angular momentum is conserved in absence of an applied torque.
(translational invariance of physical laws)
(rotational invariance of physical laws)
It’s time for some demos…..

9. L is conserved!
i
f 
 
but
f
f
i
i I
I
L 
 

The final moment of
inertia is less, so
How about the kinetic energy?
f
f
i
i
I
L
K
I
L
K
2
2
2
2


Since the inertia decreases and L stays the
same, the kinetic energy increases!
Q: Where does the force come from to do
the work necessary to increase the kinetic
energy?
A: The work is done by
the person holding the
weights!

10. Gyroscopes show change in L from applied torque
Try the bicycle wheel demo!

11. Static Equilibrium
Static equilibrium is achieved when both the NET FORCE and
NET TORQUE on a system of objects is ZERO.
Q: What relationship must hold between M1 and
M2 for static equilibrium?
A: M1 g X1 = M2 g X2

12. Static balance.
What are the forces F1 and F2?

13. Walking the plank.
How far can the cat walk safely?

14. Which mass is heavier?
1. The hammer portion.
2. The handle portion.
3. They have the same
mass.
Balance
Point
Cut at balance point

15. Find the Center of Mass
Left scale reads 290N, right scale reads 122N. Find total mass M and Rcm.

16. Find the forces.
What is the tension in the wire? What are the horizontal and
vertical components of force exerted by the bolt on the rod? Let
the mass of the rod be negligible.
Solution strategy has three steps:
1. Draw the free-body diagram
2. Write the force equations.
3. Write the torque equations.

17. Atwood Machine with Massive Pulley
Pulley with moment of inertia
I, radius R. Given M1, M2, and
H, what is the speed of M1
just before it hits the ground?
Strategy: Use conservation of mechanical energy.
Initial kinetic energy is 0. Initial potential energy is M1gH.
Final kinetic energy is translational energy of both blocks plus
rotational energy of pulley.
Final potential energy is M2gH.
Set final energy equal to initial energy.
Three steps: 1. Write down initial kinetic and potential energy.
2. Write down final kinetic and potential energy.
3. Set them equal (no friction).
HINT:

18. Static balance and a strange yo-yo.
The mass M of the yo-yo is
known. The ratio of r and R is
known. What is the tension T1
and T2, and mass m?
Strategy:
1. Write down torque equation for yo-yo.
2. Write down force equation for yo-yo
and mass m.
3. Eliminate unknowns.

19. The case of the strange yo-yo.
0
0
0
2
1
2
2
1









Mg
T
T
mg
T
R
T
r
T
Equations to solve:
Torque
Force on mass m.
Force on yo-yo.
mg
T
r
R
T
T


2
2
1
Mg
M is known, R/r is known.
Step 1: Rearrange. Step 2: Substitute.
Mg
r
R
T
Mg
T
r
R
T













1
0
2
2
2
1
1
1
1
2






a
M
m
a
a
Mg
T
a
Mg
T
Step 3: Solve (a=R/r).