Rotating on a small platform attached by massless rods to a pivot

In summary: I thought you were asking for the moment of inertia of the disk with respect to its center (which is 1/2MR^2).In summary, Bob the monkey is on a small platform that is attached by massless rods to a frictionless pivot. If only the platform were rotated a half-circle from its current location, Bob could reach the banana shown. However, being a smart monkey, he starts spinning around his center of mass on the platform, and soon has his snack. If Bob spins himself with an angular speed of 70 rad/s, how fast do the platform and Bob rotate about the frictionless pivot? It takes Bob approximately 1.5 seconds to get his snack.
  • #36
FlipStyle1308 said:
Thanks, what about the second part of the question:

How long does it take for Bob to get his snack?
You've presumably found [itex]\omega[/itex], now use it. How many radians must he traverse to reach the banana?
 
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  • #37
Radians = pi, right?
 
  • #38
FlipStyle1308 said:
Radians = pi, right?

What do you mean by that?

~H
 
  • #39
He travels a total distance of pi, which is the distance of 1/2 a circle.
 
  • #40
FlipStyle1308 said:
He travels a total distance of pi, which is the distance of 1/2 a circle.

Yeah, that's right (sorry I didn't read the Doc's post above). Yes, he must travel pi radians to reach his prize.

~H
 
  • #41
What other info do I need? Is it w initial and w final? Is the w I found the initial or final?
 
  • #42
FlipStyle1308 said:
What other info do I need? Is it w initial and w final? Is the w I found the initial or final?

The monkey's [itex]\omega[/itex] is given to you (70 rad\s). You should have found the [itex]\omega[/itex] of the monkey and the platform rotating about the central pole, (this should be negative).

~H
 
  • #43
So pi = [(70 + 0.707)/2]t ? By the way, on WebAssign, when I put in my answer of -0.707 for the first part of the question, it was wrong, but when I put 0.707 without the -, it was correct, so I guess I have to use 0.707 in this equation. But is this the right equation to use?
 
  • #44
FlipStyle1308 said:
So pi = [(70 + 0.707)/2]t But is this the right equation to use?

No. The 70 rad/s is simply how fast the platform is rotating about its own axis, not how fast the platform is rotating about the central pole. You need to use the 0.707 rad/s, this is a constant speed (because no net torques act). You can simply use speed = distance/time . Regarding the negative answer, the negative sign simply indicates that the platform is rotating about its own axis in the opposite direction to which it is rotating about the central pole. Do you understand this? I don't think I explained it very well?

~H
 
  • #45
Um...so pi = [(0.707 + pi/t)/2]t ?
 
  • #46
FlipStyle1308 said:
Um...so pi = [(0.707 + pi/t)/2]t ?

Not quite, its probably so simple your trying to overcomplicate things. As I said above;

[tex]v = \frac{ds}{dt} \Rightarrow \omega = \frac{d\theta}{dt}[/tex]

[tex]dt = \frac{\theta}{\omega}[/tex]

Where [itex]\omega[/itex] is the angle of travel in radians.

Can you go from here?

~H
 
  • #47
So w=d/t, t=d/w=pi/0.707?
 
  • #48
FlipStyle1308 said:
So w=d/t, t=d/w=pi/0.707?

That's what I'd do :smile:

~H
 
  • #49
Thank you!
 

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