Transcript Chapter 8, Part V

Problem 31
Find the moments of inertia about the x & y axes:
m = 1.8 kg, M = 3.1 kg
Problem 40
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?
Example 8-11
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 = 0.385 kg m2
Example 8-12
The same pulley, connected to
a bucket of weight mg = 15 N
(m = 1.53 kg). M = 4 kg
I = 0.385 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

Translation-Rotation Analogues & Connections
Translation
Rotation
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)
Section 8-7: 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)
Translation-Rotation Analogues & Connections
Translation
Rotation
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)
Sect. 8-7: 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
Example 8-13
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
 v = 0, ω = 0
where KE has 2 parts!!
(KE)trans = (½)Mv2
(KE)rot = (½)Iω2
v=?

y=0
Conceptual Example 8-14: 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)MR2 (also, v = ωR)
Friction: Necessary for
objects to roll without
slipping. Example: work
done by friction hasn’t
been included (used
KE +PE =const). WHY?
Because Ffr  Motion
 Ffr does no work!