StarvingAutist
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Apparently, the equations of motion of this system should result in SHM if the small-angle approximation is used. However, I get an angular velocity term in the theta equation.. what am I not seeing?

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You'll have to rotate it; it's not behaving.
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lerjj
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(Original post by StarvingAutist)
Apparently, the equations of motion of this system should result in SHM if the small-angle approximation is used. However, I get an angular velocity term in the theta equation.. what am I not seeing?

Name:  2016-05-02 18.36.46.jpg
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You'll have to rotate it; it's not behaving.
Is the whole thing simply moving to the right at constant velocity?
Can you not just change reference frame to where it's stationary and then show it's SHM there, and then in the moving frame it will have a translation added on top of that?
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StarvingAutist
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(Original post by lerjj)
Is the whole thing simply moving to the right at constant velocity?
Can you not just change reference frame to where it's stationary and then show it's SHM there, and then in the moving frame it will have a translation added on top of that?
No, the support also performs SHM; it moves in the opposite direction to the pendulum and starts doing so when the pendulum is released. I don't think I could change to the CoM frame because the CoM velocity wouldn't be constant.
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lerjj
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(Original post by StarvingAutist)
No, the support also performs SHM; it moves in the opposite direction to the pendulum and starts doing so when the pendulum is released. I don't think I could change to the CoM frame because the CoM velocity wouldn't be constant.
Okay, sorry, couldn't read the question clearly due to the rotation thing. So basically, you have a pendulum and the thing it's hanging from is oscillating horizontally?
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StarvingAutist
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(Original post by lerjj)
Okay, sorry, couldn't read the question clearly due to the rotation thing. So basically, you have a pendulum and the thing it's hanging from is oscillating horizontally?
Yeah basically but I have to show that it does; it's not a given
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atsruser
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(Original post by StarvingAutist)
Apparently, the equations of motion of this system should result in SHM if the small-angle approximation is used. However, I get an angular velocity term in the theta equation.. what am I not seeing?

Name:  2016-05-02 18.36.46.jpg
Views: 645
Size:  172.0 KB

You'll have to rotate it; it's not behaving.
I'm afraid I don't have the time to look at this at the moment, but you have taken a Lagrangian approach for some reason. However, it would seem to be more straightforward to consider conservation of momentum.

If you imagine the pendulum hanging vertically at rest, and then given an impulse to the right by someone inside the trolley, then by conservation of momentum, the trolley must move to the left in such a way as to keep the c-o-m of the system stationary. You then need to check that SHM for small pendulum angles => SHM for the c-o-m of the trolley.

Or is that too simple? Maybe I've misunderstood the problem.
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The-Spartan
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(Original post by StarvingAutist)
Apparently, the equations of motion of this system should result in SHM if the small-angle approximation is used. However, I get an angular velocity term in the theta equation.. what am I not seeing?

Name:  2016-05-02 18.36.46.jpg
Views: 645
Size:  172.0 KB

You'll have to rotate it; it's not behaving.
(Original post by atsruser)
I'm afraid I don't have the time to look at this at the moment, but you have taken a Lagrangian approach for some reason. However, it would seem to be more straightforward to consider conservation of momentum.

If you imagine the pendulum hanging vertically at rest, and then given an impulse to the right by someone inside the trolley, then by conservation of momentum, the trolley must move to the left in such a way as to keep the c-o-m of the system stationary. You then need to check that SHM for small pendulum angles => SHM for the c-o-m of the trolley.

Or is that too simple? Maybe I've misunderstood the problem.
Just for reference, there is a great paper done by Gabriela Gonzalez on this matter here
Should clear things up a little bit.
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atsruser
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(Original post by StarvingAutist)
Apparently, the equations of motion of this system should result in SHM if the small-angle approximation is used. However, I get an angular velocity term in the theta equation.. what am I not seeing?

Name:  2016-05-02 18.36.46.jpg
Views: 645
Size:  172.0 KB

You'll have to rotate it; it's not behaving.
Addendum to previous response: I've though about this now when not tired, and I think that it's trickier than I suggested - I was thinking of taking a small angle pendulum which moves with SHM, then analysing the coupled system to show that the other bit also had SHM. However, since the support moves, you can't assume that the pendulum does, in fact, move with SHM. So you'll need a more sophisticated approach. Maybe the referenced paper will help.
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lerjj
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(Original post by atsruser)
I'm afraid I don't have the time to look at this at the moment, but you have taken a Lagrangian approach for some reason. However, it would seem to be more straightforward to consider conservation of momentum.

If you imagine the pendulum hanging vertically at rest, and then given an impulse to the right by someone inside the trolley, then by conservation of momentum, the trolley must move to the left in such a way as to keep the c-o-m of the system stationary. You then need to check that SHM for small pendulum angles => SHM for the c-o-m of the trolley.

Or is that too simple? Maybe I've misunderstood the problem.
Pendulums aren't closed systems though since they're subject to gravity. Doesn't that rule out a conservation of momentum argument?
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atsruser
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(Original post by lerjj)
Pendulums aren't closed systems though since they're subject to gravity. Doesn't that rule out a conservation of momentum argument?
The momentum of the pendulum itself is not conserved, but it is conserved for the whole pendulum-trolley system, since there is no external horizontal force acting => horizontal momentum is conserved, and there is no nett vertical force acting (though the reaction of the ground on the trolley is time dependent) => vertical momentum is conserved. The c-o-m of the whole system remains stationary while the trolley rolls back and forth a bit.

But I think that it's best just to slog it out with the full Lagrangian approach for this problem.
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