Download Massive Pulleys Review

Massive Pulleys Review
Massive pulleys have rotational inertia! It takes a net
torque to change their angular velocity ω .
τ net = Iα
Torque on a pulley is caused by the rope tension T.
If there is no slipping, its magnitude is T times the
radius the rope goes around the pulley.
2
kg
In solving massive pulley problems you can relate linear
motion of rope to angular motion of pulley.
Clicker question -2
Set frequency to BA
A mass m hangs from string wrapped around a pulley of
radius R. The pulley has a moment of inertia I and its pivot is
frictionless. Because of gravity the mass falls and the pulley
rotates. The magnitude of the torque on the pulley is…
A.  greater than mgR
B.  less than mgR
C.  equal to mgR
T
R
FG = mg
(Hint: Is the tension in the string = mg?)
Free body diagram of mass gives T=mg-ma
when a is positive so T<mg so τ = rT < Rmg
m
2
Clicker question -1
Set frequency to BA
For the 2 kg weight attached to 6 kg pulley with radius
of 0.10 m, what is the kinetic energy of the weight
after dropping for 1 second. Remember a=4 m/s2 and
K=½mv2, K=½Iω2,and Idisk=½MR2.
A.  2 J
B.  4 J
C.  8 J
D.  16 J
E.  32 J
After 1 second
v = v0 + at = 0 + 4 m/s2 ⋅1 s = 4 m/s
2
kg
Kinetic energy for the weight is
2
2
1
1
K = 2 mv = 2 2 kg ⋅ ( 4 m/s) = 16 J
3
Clicker question 0
Set frequency to BA
For 2 kg weight attached to 6 kg pulley with R=0.10 m,
what is the kinetic energy of the pulley after the weight
drops for 1 second. Remember v=4 m/s for the weight
after 1 second and K=½mv2 and K=½Iω2 and Idisk=½MR2.
A.  4 J
B.  8 J
C.  16 J
D.  24 J
E.  48 J
The weight velocity and the outside
pulley velocity are the same since
they are connected by the rope.
2
kg
Can get angular velocity from rim velocity
v
and then find kinetic energy
ω=
R
2
2
2
2 ! v $
2
1
1 1
1
1
K = 2 Iω = 2 ( 2 MR ) # & = 4 Mv = 4 6 kg ⋅ ( 4 m/s) = 24 J
"R%
4
Massive pulleys
A 2 kg weight is attached to the end of a rope coiled
around a pulley with mass of 6 kg and radius of 0.1 m.
The acceleration of the weight is found to be 4 m/s2.
After 1 second the total kinetic energy was found to be
K total = K weight + K pulley = 16 J + 24 J = 40 J
2
kg
After 1 second the weight will descend
Δy = v0 y t + 12 ay t 2 = 0 + 12 ⋅ 4 m/s2 ⋅ (1 s)2 = 2 m
losing gravitational potential energy
Ugrav = mgh = 2 kg ⋅10 m/s2 ⋅ 2 m = 40 J
Conservation of
energy still holds!
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Clicker question 1
Set frequency to BA
Two unequal weights are attached to a rope that is
looped around a pulley. The rope does not slide with
respect to the pulley. The pulley is massive and has a
frictionless pivot. After the blocks are released, what
T2
T1
is the relationship between T1 and T2?
2
1kg
A.  T1 > T2
There is a related CAPA
kg
B.  T1 = T2
question due next week.
C.  T1 < T2
D.  Impossible to tell
What are the torques acting on the
pulley and what is the net torque?
Consider just
τ net = RT2 − RT1 = R (T2 − T1 )
the pulley:
R
T2
T1
From Newton’s 2nd law: τ net = Iα
Since the pulley will start rotating counter-clockwise,
there must be positive angular acceleration so T2 > T1.
Possible pitfall
Sometimes more than one R in the problem.
R is found in several places (choose correctly)
1. Torque calculation τ = rF sin φ
2
1
I
=
MR
2. Moment of inertia calculation disk 2
2 kg
3. Conversion between rope linear velocity
and pulley angular velocity v = Rω or rope
linear acceleration and pulley angular
acceleration a = Rα
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Very important notes on gravity
The potential energy due to gravity for an extended
body is measured from the center of mass.
When calculating torque due to gravity, the location of the
force of gravity can be assumed to be the center of mass.
U grav = Mgh = Mg L sin 30° τ = L Mg sin 60°
2
2
Note the torque will change
as the rod falls due to the
angle changing
L
M
h = L2 sin 30°
FG = Mg
30°
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A few more tips
Be careful when relating signs of angular acceleration
and linear (tangential) acceleration or angular velocity
and linear velocity
If torques and forces aren’t helping, try
conservation of energy (and vice versa)
Everything you have already learned continues to work
Newton’s 2nd law is always true
(for inertial reference frames)
Conservation of energy still applies.
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Time for a Torque Wave
10
Rotation & Translation
Problems so far had a stationary axis.
Other problems have a moving axis like
a boomerang, bowling ball, or yo-yo.
This is a combination of translational
motion and rotational motion
Split motion into two parts
1. Translational motion of the center of mass
2. Rotational motion around the center of mass
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Separate kinetic energy terms
Separate kinetic energy into translational and
2
rotational parts
K = KCM + Krot = 12 MvCM
+ 12 ICMω 2
Sometimes center of mass velocity & angular velocity are related.
Rolling without slipping:
vCM = Rω
Note this requires static friction to prevent slipping
rotational
translational
vCM
+
ω
total
=
vCM
12
Clicker question 2
Set frequency to BA
A sphere, hoop, and cylinder, all with the same mass M and
radius R, are rolling along with the same speed v. Which one
has the most kinetic energy?
v
v
v
A.  Sphere
B.  Hoop
C.  Disk
D.  All have the same kinetic energy
2
1
K
=
Mv
All have the same translational kinetic energy: CM 2
CM
All have the same angular velocity: ω = vCM / R
The hoop has the largest moment of inertia so
2
1
K
=
I
ω
it will have the most rotational kinetic energy:
rot
2 CM
So hoop has the most kinetic energy: K = K CM + K13rot
Hoop vs Disk
A hoop, and disk, with the same mass M and radius R, start
rolling down at ramp, starting from rest. Which one reaches the
bottom first?
h
Ug = Mgh
1
2
2
CM
K = Mv
1
2
+ I CMω
2
vCM = Rω
2
1
1 ! vCM $ 1 2
I
2
K = MvCM + ( I ) #
& = vCM (M + 2 ) = Mgh
2
2 " R % 2
R
vCM =
2Mgh
I
M+ 2
R
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Yo-yo
A yo-yo of mass M and radius R starts from rest
and is dropped, attached by an unrolling string.
What is the center of mass velocity after falling a height h?
Asking for velocity suggests conservation of energy.
We know that falling a height loses gravitational
potential energy which in this case is Ug = Mgh
1
2
2
CM
This energy goes into kinetic energy: K = Mv
Since this is rolling without slipping
1
2
+ I CMω
vCM = Rω
2
1
1!1
3
2
2 $ ! vCM $
2
Setting Ug = K
K = MvCM + # MR &#
=
Mv
&
CM
%" R % 4
2
2"2
4
vCM =
gh
3
15
2
Pulling the yo-yo
Yo-yo with inner radius R1 and outer radius R2 is
on a table. String is pulled with force F as shown.
There is enough static friction to prevent the yo-yo
from sliding. Which way does the yo-yo move?
F
Given forces and asking direction suggests Newton’s 2nd law.
τ net = R2 Ff − R1F = I CMα
Fnet = F − Ff = MaCM
F
Ff
Rolling without slipping: vCM = Rω and aCM = Rα
but you need to make sure you get the signs right!
1
1
2 aCM
= MR2 aCM
so R2 Ff − R1F = MR2
2
R2 2
1
R
F
−
Ma
−
R
F
=
MR2 aCM
gives
F
=
F
−
Ma
Substituting f
2(
CM )
1
CM
2
Rearranging: R2 F − R1F = 23 MR2 aCM
2 ( R2 − R1 ) F
aCM =
3MR2
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Pulling the yo-yo
Using the extended free body diagram and
Newton’s 2nd law we found the equations:
Fnet = F − Ff = MaCM
R1
R2
F
Ff
τ net = R2 Ff − R1F = I CMα
Applying rolling without slipping: vCM = Rω and
aCM = Rα
2 ( R2 − R1 ) F If R2 > R1 then aCM > 0
After some math: aCM =
3MR2
(which is to the left)
If R1 = 0 then F exerts no torque and aCM = 2F
3M
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Next Week – Chapter 11
Angular Momentum!!!
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