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Announcements• CAPA #11 due this Friday at 10 pm
• Reading: Finish Chapter 8, Start Chapter 9.1-9.4
• Section – this week Lab #4: Rotations
• Midterm Exam #3 on Tuesday November 8th, 2011 details given in class on Wednesday practice exam and solutions on CULearn formula sheet to be posted on web page
• Fraction of all clicker questions answered posted on CULearn. Email me with your clicker ID, name, student ID if you believe it is incorrect.
I miri2
i
Which has a larger moment of inertia?
A) IA > IB B) IA < IB
C) IA = IB D) Impossible to tell.
IB 2mL2 IA 2(2m)L
2
2
mL2
Clicker Question Room Frequency BA
Consider two masses each of size 2m at the ends of a light rod of length L with an axis of rotation through the center of the
rod. The rod is doubled in length and the masses are halved.
A bar has four forces, all of the same magnitude, exerted on it, as shown.
What is the sign of the net torque about the axis of rotation? Use the sign convention shown.
Clicker Question Room Frequency BA
A) torque is zero B) positive (+) C) negative (–)
tnet = + (F)(L) + (F)(L/2) + (F)(L/2) – (F)(L) = +FL
Rotational Kinetic EnergyDoes this object have
translational kinetic energy?
No, zero net translational velocity of the object.
However, there is motion of each piece of the object and thus there must be kinetic energy.
Each piece of the donut has a velocity v = w r.
KE = ½ mv2 = ½ m (w r)2
KE = ½ I w2 Rotational KE
Rolling Kinetic Energy
Translation Rotation
KE (total) = KE (translation) + KE (rotation)
KEtotal = ½ mv2 + ½ I w2
Both pieces in units of Joules.
* Rolling without slipping means v = w r.One revolution Dq=2p leads to displacement of 2pr
KEtot 1
2Mv2
1
2I 2
Isphere 2
5MR2
Ihoop MR2
Idisk 1
2MR2
Which object has the largest total
kinetic energy at the bottom of the ramp?
A) Sphere B) Disk C) Hoop D) All the same.
Clicker Question Room Frequency BA
ffii PEKEPEKE 00 fKEMgH
MgHKE f
M
All have the same total KE.
KEtot 1
2Mv2
1
2I 2
Isphere 2
5MR2
Ihoop MR2
Idisk 1
2MR2
M
ffii PEKEPEKE
02
1
2
10 22
IMvMgH
2
22
2
1
2
1
R
vIMvMgH
22
2
1
R
IMvMgH 2MRI
MgHRv
KEtot 1
2Mv2
1
2I 2
Isphere 2
5MR2
Ihoop MR2
Idisk 1
2MR2
M
2MRI
MgHRv
gHMRMR
MgHRv
7
10
5
2 22
Sphere
KEtot 1
2Mv2
1
2I 2
Isphere 2
5MR2
Ihoop MR2
Idisk 1
2MR2
vsphere 10
7gH
Which has the greater speed at the bottom of the ramp, the sphere that rolls down the ramp or a block of the same mass
that slides down the ramp?
(Assume sliding friction is negligible)
A) Block B) Sphere C) Both the same
Clicker Question Room Frequency BA
ffii PEKEPEKE
02
10 2 MvMgH
Block gHv 2
KEtot 1
2Mv2
1
2I 2
I sphere 2
5MR2
Ihoop MR2
Idisk 1
2MR2
Sphere:
vhoop gH
Hoop:
vdisk 4
3gH
Disk:
Who wins the race to the bottom…… sphere, disk, hoop?
2MRI
MgHRv
gHv7
10
Smallest moment of inertia Iwill have the largest translational
velocity at the bottom.
Demonstration
Which object will go furthest up the incline?
A) Puck B) Disk C) Hoop D) Same height.
Ihoop MR2 , Idisk 1
2MR2
The hoop has the largest moment of inertia, and
therefore thehighest total kinetic
energy.
H
Clicker Question Room Frequency BA
ffii PEKEPEKE
MgHr
vIMv
22
2
1
2
1
Recall:
Momentum p = mv L = Iω Angular momentum
Relation to force F = Δp/Δt τ = ΔL/Δt Relation to torque
No external force Δp = 0 ΔL = 0 No external torque (momentum is conserved) (angular momentum is conserved)
Ii ωi = If ωf
Conservation of Angular Momentum
Li = Lf
L I
Lt
Ii i = I f f
(if Fext 0)
Ii largeωi small
Ii smallωi large
By changing the distribution of mass, the moment of inertia is changed.
By conservation of angular momentum, the angular velocity is therefore modified.
Conservation of L:
finalinitial LL
ffii II
if
if I
I
I1 largeω1 small
By changing the distribution of mass, the moment of inertia is changed.
By conservation of angular momentum, the angular velocity is therefore modified.
Conservation of L:
I2 smallω2 large
I3 largeω3 small
L I
Lt
Ii i = I f f
(if Fext 0)
finalinitial LL
ffii II
if
if I
I
Hoberman Sphere