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Moment of inertia of a spinning disc pendulum

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1. Homework Statement
Find the moment of intertia of a pendulum, consisting of a disc free to spin attached to a rod that is hinged at one end.


2. Homework Equations
Moment of intertia of rod hinged at end = (1/3)Ml2
Moment of intertia of disc = (1/2)mR2 + ml2

3. The Attempt at a Solution
Why does the answer disregard the part moment of intertia of the disc (1/2)mR2 that is spinning on its own axis, and only taking into account the ml2?

Here's what the answer wrote:

"If the disk is not fixed to the rod, then it will not rotate as the pendulum oscillates.
Therefore it does not contribute to the moment of inertia. Notice that the pendulum is no
longer a rigid body. So the total moment of inertia is only due to the rod and the disk
treated as a point like object."

I got completely lost by the first sentence. Why does the disc not contribute to the moment of inertia when it is spinning? I thought the idea behind moment of inertia is linked to the rotational kinetic energy it possesses? Oscillating a spinning disc does not rob it of its kinetic energy!
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
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Well the disk is spinning, but along a different axis. It's spin does not contribute to the rotational inertia calculated about the axis of the pendulum's hinge. It would be no easier or harder to stop the pendulum from rotating if the disk were free to spin, or if it were rigidly attached. So this suggests that its rotational inertia about its own axis of rotation should not be contributing to the rotational inertia of the entire system.

Now if you were asked to calculate the total rotational kinetic energy of the system, you would have to include the spin of the disk about its axis.

In my opinion, this is a sort of dumb pointlessly confusing question. Like something you would find on a chemistry exam :D
 
Last edited:
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I still dont understand! help?
 
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Actually i've just realized that what I said is a bit incorrect. The disk will not spin. And actually if it were rigidly attached, then the extra term would have to be included. Sorry to be misleading.
The idea is that if the disk is free to spin, then it actually will not spin on its own axis. If you imagine an arrow painted on the disk pointing directly upwards when the pendulum is at maximum displacement, then the arrow will continue to point directly up. However if the disk were rigidly attached, if you can imagine an arrow pointing to the axis of the pendulum when it is at maximum displacement, then the arrow will always point to the axis. We can see in the second case that the disk is actually spinning a bit on its own axis, but in the first case it is not.
Does this help at all?
 
283
0
Well the disk is spinning, but along a different axis. It's spin does not contribute to the rotational inertia calculated about the axis of the pendulum's hinge. It would be no easier or harder to stop the pendulum from rotating if the disk were free to spin, or if it were rigidly attached. So this suggests that its rotational inertia about its own axis of rotation should not be contributing to the rotational inertia of the entire system.
Wow actually this whole paragraph is garbage. Please disregard it. I hope that I haven't confused you.
 
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lol it's okay, I appreciate your effort! Any senior contributors around here mind to help?
 

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