1. Rotational Dynamics Continued

2. Find the moments of inertia about the x & y axes:
Problem 31
Find the moments of inertia about the x & y axes:
m = 1.8 kg, M = 3.1 kg

3. Problem m each blade = 160 kg. Moment of inertia I? Starts from
rest, torque τ to get ω = 5 rev/s (10π rad/s) in t = 8 s?

4. Example M = 4 kg, FT = 15 N Frictional torque: τfr = 1.1 m N
t = 0, ω0 = 0; t = 3 s, ω = 30 rad/s
I = ?
Calculate the angular acceleration:
ω = ω0+ αt, α = (ω - ω0)/t
= 10 rad/s2
N’s 2nd Law: ∑τ = Iα
FTR - τfr = Iα
 I = [(15)(0.33) -1.1]/10
I = kg m2

5. a  Example 8-12 a) α = ? (pulley) b) t = 0, at rest.
The same pulley, connected to
a bucket of weight mg = 15 N
(m = 1.53 kg). M = 4 kg
I = kg m2; τfr = 1.1 m N
a) α = ? (pulley)
a = ? (bucket)
b) t = 0, at rest.
t = 3 s, ω = ? (pulley)
v = ? (bucket) a 

6. Translation-Rotation Analogues & Connections
Displacement x θ
Velocity v ω
Acceleration a α
Force (Torque) F τ
Mass (moment of inertia) m I
Newton’s 2nd Law: ∑F = ma ∑τ = Iα
Kinetic Energy (KE) (½)mv2 ?
CONNECTIONS
v = rω, atan= rα, aR = (v2/r) = ω2 r
τ = rF, I = ∑(mr2)

7. Rotational Kinetic Energy
Translational motion (Ch. 6): (KE)trans = (½)mv2
Rigid body rotation, angular velocity ω. Rigid
 Every point has the same ω. Body is made of particles, masses m.
For each m at a distance r from the rotation axis:
v = rω. The Rotational KE is:
(KE)rot = ∑[(½)mv2] = (½)∑(mr2ω2) = (½)∑(mr2)ω2
ω2 goes outside the sum, since it’s the same everywhere in the body
As we just saw, the moment of inertia, I  ∑(mr2)
 (KE)rot = (½)Iω2
(Analogous to (½)mv2)

8. Translation-Rotation Analogues & Connections
Displacement x θ
Velocity v ω
Acceleration a α
Force (Torque) F τ
Mass (moment of inertia) m I
Newton’s 2nd Law ∑F = ma ∑τ = Iα
Kinetic Energy (KE) (½)mv2 (½)Iω2
CONNECTIONS
v = rω, atan= rα, aR = (v2/r) = ω2 r
τ = rF, I = ∑(mr2)

9. Rotational + Translational KE
Rigid body rotation: (KE)rot = (½)Iω2
Now, consider a rigid body, mass M, rotating (angular velocity ω) about an axis through the CM. At the same time, the CM is translating with velocity vCM
Example, a wheel rolling without friction. For this, we saw earlier that vCM = rω.
The KE now has 2 parts: (KE)trans & (KE)rot
 Total KE = translational KE + rotational KE
KE = (KE)trans + (KE)rot
or KE = (½)M(vCM)2 + (½)ICMω2
where: ICM = Moment of inertia about an axis through the CM

10. Example  y = 0 KE+PE conservation: where KE has 2 parts!!
A sphere rolls down an
incline (no slipping or sliding).
KE+PE conservation:
(½)Mv2 + (½)Iω2 + MgH = constant, or
(KE)1 +(PE)1 = (KE)2 + (PE)2
where KE has 2 parts!!
(KE)trans = (½)Mv2
(KE)rot = (½)Iω2
 v = 0, ω = 0
v = ? 
 y = 0

11. Conceptual Example: Who Wins the Race?
v increases as I
decreases! Demonstration!
MgH = (½)Mv2 + (½)ICMω2
Gravitational PE is Converted to
Translational + Rotational KE!
Hoop: ICM = MR2 Cylinder: ICM = (½)MR2
Sphere: ICM = (2/5)MR (also, v = ωR)

12. Friction is necessary for objects to roll without slipping
Friction is necessary for objects to roll without slipping. In the example, the work done by friction hasn’t been included (we used KE +PE = const). WHY? Because the friction force Ffr is  to the motion direction so it does no work!