Spinning Bicyle Wheel, Person, and Platform


by e(ho0n3
Tags: bicyle, person, platform, spinning, wheel
e(ho0n3
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#1
Jun12-04, 03:51 PM
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A person stadns on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is [itex]I_p[/itex]. The person holds a spinning bicycle wheel with axis horizontal. The wheel has momemnt of inertia [itex]I_w[/itex] and angular velocity [itex]\omega_w[/itex]. What will the angular velocity [itex]\omega_p[/itex] of the platform if the person moves the axis of the wheel so that it points vertically upward.

Using conservation of momentum:
[tex]\vec{L}_w = \vec{L'}_w + \vec{L}_p[/tex]
where
[tex]\begin{align*}
\vec{L}_w &= (I_w\omega_w, 0)\\
\vec{L'}_w &= (0, I_w\omega_w)\\
\vec{L}_p &= I_p\vec{\omega}_p = I_p(\omega_1, \omega_2)
\end{align}[/tex]
Solving for [itex]\omega_1[/itex] and [itex]\omega_2[/itex] gives
[tex]\begin{align*}
\omega_1 &= \frac{I_w\omega_w}{I_p}\\
\omega_2 &= -\frac{I_w\omega_w}{I_p}
\end{align}[/tex]
And since [itex]\omega_p = \sqrt{\omega_1^2 + \omega_2^2}[/itex] then
[tex]\omega_p = \frac{\sqrt{2}I_w\omega_w}{I_p}[/tex]
According to my book though,
[tex]\omega_p = \frac{I_w\omega_w}{I_p}[/tex]
Is it just me, or is my book on crack?
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e(ho0n3
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#2
Jun12-04, 07:10 PM
P: 1,370
Quote Quote by e(ho0n3
Is it just me, or is my book on crack?
Or is that the other way around!?
Doc Al
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#3
Jun12-04, 07:12 PM
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The book seems correct to me. Note that angular momentum is only conserved about the vertical axis: the platform is constrained and only freely rotates about that vertical axis.

So: initial angular momentum about vertical axis is zero! The wheel gains [itex]+ I_w\omega_w[/itex], so the platform must gain [itex]- I_w\omega_w[/itex]

gnome
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#4
Jun12-04, 07:18 PM
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Spinning Bicyle Wheel, Person, and Platform


The book's answer looks OK to me.

Initially, the angular momentum about the vertical axis is 0, so conservation of momentum requires that this remains true.

When the wheel's axis turns vertical, the wheel now has angular momentum about the vertical axis equal in magnitude to it's initial angular momentum, Lw = Iwωw.

The (person+wheel) must have angular momentum of equal magnitude in the opposite direction in order for the net total vertical angular momentum to remain 0.

Therefore,
[tex]I_p\omega_p = I_w\omega_w[/tex]
[tex]\omega_p = \frac{I_w\omega_w}{I_p}[/tex]
gnome
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#5
Jun12-04, 07:20 PM
P: 1,047
Yeah, like he said...
e(ho0n3
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#6
Jun12-04, 08:33 PM
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What!? I didn't know angular momentum was picky about the axis it chooses to remain conservative in. Now I'm really confused. So how do I determine the axis where angular momentum is conserved in general when I have a system of rotating objects?
gnome
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#7
Jun12-04, 10:54 PM
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Momentum, whether linear or angular, is a vector so conservation requires that you consider the direction as well as the magnitude.

As far as which axis...that's determined by the physical characteristics of the system you're looking at. As the wheel ends up spinning on a vertical axis, and the man is on a platform that spins freely on a vertical axis, the vertical is the axis you have to worry about.

The next question is, what happened to the momentum about the horizontal axis? Why does that seem not to have been conserved? If you think a bit about that aspect, you should come up with the answer.
e(ho0n3
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#8
Jun12-04, 11:44 PM
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Quote Quote by gnome
The next question is, what happened to the momentum about the horizontal axis? Why does that seem not to have been conserved? If you think a bit about that aspect, you should come up with the answer.
Angular momemtum is conserved only when the net external torque on the system is zero according to my understanding. The only thing I can think of is that the person causing the wheel to change its rotation axis has applied a net external torque, hence causing momentum not to be conserved. But this is a contradiction! I still don't get it. I think my error is that I'm thinking of angular momentum like linear momentum. You'll have to forgive me as my physical intuition sucks.
Doc Al
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#9
Jun13-04, 07:09 AM
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Quote Quote by e(ho0n3
Angular momemtum is conserved only when the net external torque on the system is zero according to my understanding.
Right! And depending on how you define your system, the net external torque may be zero about one axis but nonzero about another.
The only thing I can think of is that the person causing the wheel to change its rotation axis has applied a net external torque, hence causing momentum not to be conserved.
Keep going. It's not just the person/platform and wheel floating out in space. The platform is attached to the earth.
But this is a contradiction!
Why? Since nothing can exert an external torque about the vertical axis, then the angular momentum of the person/platform plus wheel must be conserved about that direction.

Of course, if you include the earth as part of your system--then angular momentum is conserved in all directions!
I think my error is that I'm thinking of angular momentum like linear momentum.
Not sure what you mean by that, but it is definitely true that angular momentum can be a much subtler concept. But they are both vectors. You just rarely get a (simple) problem with linear momentum where it's only conserved in one direction.
arildno
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#10
Jun13-04, 09:11 AM
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Just adding a few comments to what Doc Al and gnome have said:

Consider the system consisting of platform, person and wheel.
The platform has an axle through it (fixed to the ground), about which it may rotate freely.
(The "axle" also has a "plate" attached to it, which the platform rests upon; there is no friction between the platform or the "plate" either.)


Hence, no external forces exist which may produce torques about the axis
(These would have been frictional forces between platform and axle)

Therefore, in the interval (0,T) between initial state and final state, the moment-of momentum equation, with the torques computed relative to the C.M of the system, we have:
[tex]\vec{\tau}=\frac{d\vec{S}_{C.M}}{dt},\vec{\tau}\cdot\vec{k}=0[/tex]

Here, [tex]\vec{k}[/tex] is parallell to the axle, [tex]\vec{S}_{C.M}[/tex] is the angular momentum of the system with respect to C.M, while
[tex]\vec{\tau}[/tex] is the net torque of external forces acting on the system.

Clearly, we therefore have:
[tex]\vec{S}_{C.M}(0)\cdot\vec{k}=\vec{S}_{C.M}(T)\cdot\vec{k}[/tex]

This gives the relation in the book by identifying
[tex]\vec{S}_{C.M}(0)\cdot\vec{k}=0[/tex]
arildno
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#11
Jun13-04, 10:37 AM
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Note:
The external, torque-producing forces are normal forces from the axle onto the platform .
In reality, this will cause the platform to wobble slightly between initial state and final state.

The force responsible for the platform's rotation, is the frictional force between the person's feet and the platform.


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