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

[FlipStyle1308]

I recall my teacher explaining this concept in class, but he did not go over how we would solve these types of problems:

Bob the monkey (20 kg) is on a small platform (2 kg), which is attached by massless rods (R = 3m) to a frictionless pivot, and initially at rest.
http://img59.imageshack.us/img59/5623/bobthemonkey9go.jpg
Bob would like to get the banana shown, which he could reach-- if only the platform were rotated a half-circle from its current location. Being a smart monkey, he starts spinning around his center of mass on the platform, and soon has his snack.

a. What is Bob's rotational inertia about his center of mass? (You make approximate him as a sphere of radius 0.5 m).
b. What is the rotational inertia of the system around the frictionless pivot? (Remember that Bob also moves with the platform. In this calculation, treat him as a point mass.)
c. If Bob spins himself with ana ngular speed of 70 rad/s, how fast do Bob+platform rotate about the frictionless pivot? How long does it take for Bob to get his snack?

For A, I used I = (20)(2.5)^2 and (22)(2.5)^2, and both answers were wrong. I don't know what else I can use to figure out the rotational inertia around Bob's center of mass.

For B, I tried I = (2)(3)^2, but this answer was wrong.

I haven't tried C, since I felt that I would again get a wrong answer, since the previous 2 parts are incorrect. Any help on this problem would be greatly appreciated! Thanks!

 

[nrqed]

I recall my teacher explaining this concept in class, but he did not go over how we would solve these types of problems:

Bob the monkey (20 kg) is on a small platform (2 kg), which is attached by massless rods (R = 3m) to a frictionless pivot, and initially at rest.
http://img59.imageshack.us/img59/5623/bobthemonkey9go.jpg
Bob would like to get the banana shown, which he could reach-- if only the platform were rotated a half-circle from its current location. Being a smart monkey, he starts spinning around his center of mass on the platform, and soon has his snack.

a. What is Bob's rotational inertia about his center of mass? (You make approximate him as a sphere of radius 0.5 m).
b. What is the rotational inertia of the system around the frictionless pivot? (Remember that Bob also moves with the platform. In this calculation, treat him as a point mass.)
c. If Bob spins himself with ana ngular speed of 70 rad/s, how fast do Bob+platform rotate about the frictionless pivot? How long does it take for Bob to get his snack?

For A, I used I = (20)(2.5)^2 and (22)(2.5)^2, and both answers were wrong. I don't know what else I can use to figure out the rotational inertia around Bob's center of mass.

That's a fun problem! And a smart monkey!

You need the moment of inertia of a *sphere* (with respect to an axis going through its center)! Do you know the expression?

For B, I tried I = (2)(3)^2, but this answer was wrong.

why 2??
You will have 20 kg * (3 m)^2 plus the moment of inertia of the platform (you will need the moment of inertia of a full disk with respect to its center)

 

[FlipStyle1308]

So I = 7/5MR^2?

 

[nrqed]

So I = 7/5MR^2?

?? I thought it was 2/5 MR^2 for a sphere!

 

[FlipStyle1308]

Okay, for that I got 2 for Bob's rotational inertia about his center of mass.

I then used (20)(3)^2 + 2, and got 182, but got the question wrong. Did I do something wrong?

 

[nrqed]

Okay, for that I got 2 for Bob's rotational inertia about his center of mass.

I then used (20)(3)^2 + 2, and got 182, but got the question wrong. Did I do something wrong?

Ok for Bob's moment of inertia.

But in the next calculation, where did you get 2?? The 20*3^2 is the moment of inertia of Bob (now treated as a point particle) . To that you must add the moment of inertia of the disk. Do you know the formula for the I of a disk rotated with respect to its axis? Compute that (using the mass of the disk and its radius) and add that to the I of Bob (the 20*3^2).

 

[FlipStyle1308]

So I will use (20)(3)^2 + (0.5)(2)(3)?

 

[nrqed]

So I will use (20)(3)^2 + (0.5)(2)(3)?

well, it is 1/2 M R^2 (so it is squared).

But wait.. are you given the radius of the disk?? If not, we are stuck. (unless one treats the disk as having a negligible radius, in which case it contributes nothing to the total moment of inertia).

So, are you sure you gave all the information provided??

 

[FlipStyle1308]

There is no radius for the disk.

 

[nrqed]

There is no radius for the disk.

oh, my mistake! Sorry!
When I saw ''2k of the platform'' I thought it was the mass of the swivel! Sorry!

Well, then the platform can be treated as point mass also so you would add to the I of the monkey 2kg *(3 m)^2. (or you could have combined the monkey and the platform in one object of 22k and use (22kg) (3m)^2

my apologies..

 

[FlipStyle1308]

So (22)(3)^2 = 198?

 

[nrqed]

So (22)(3)^2 = 198?

yes..don't forget to include the units

 

[FlipStyle1308]

Okay cool, and what about the final part?

 

[FlipStyle1308]

Bump! Anyone able to help me finish off this problem?

 

[Hootenanny]

I think C can be solved if you consider conservation of angular momentum.

~H

 

[FlipStyle1308]

How so? Would you be able to elaborate more on that?

 

[Doc Al]

Since no external torques act about the vertical axis, the total angular momentum of the monkey-platform system about that axis remains constant.

 

[FlipStyle1308]

Okay, and how would I be able to incorporate this into what I have now...how should I set my equation up?

 

[Doc Al]

Find the total angular momentum: The monkey rotates about his axis, what's his angular momentum? The platform/monkey rotates about the central axis, what's its angular momentum?

What's the total angular momentum?

 

[FlipStyle1308]

What is the equation for total angular momentum? Is it L=Iw? I don't know L or w, so I have 2 unknowns?

 

[Hootenanny]

What is the equation for total angular momentum? Is it L=Iw? I don't know L or w, so I have 2 unknowns?

You have bob's angular velocity (w) and his moment of inerta, this will allow to calculate the angular momentum of the system.

~H

 

[FlipStyle1308]

So I am just looking for Bob's angular momentum, and not the angular momentum of the whole system?

 

[Doc Al]

You need to consider the total angular momentum, which is the sum of the Bob's internal angular momentum (about his axis) plus the angular momentum of the system about the central axis (which is what you want to find). What's the total angular momentum? (That's trivial, if you understand what's going on.)

 

[FlipStyle1308]

So I have this... L = Iw + Iw = (2)(70) + (198)w...I have 2 unknowns...the total angular momentum and the angular speed about the frictionless pivot (which I am solving for). How do I get L?

 

[Doc Al]

You should only have one unknown: You should know the total L. Hint: Before Bob gets the desire for that banana, what is L? It hasn't changed.

 

[FlipStyle1308]

Am I using L=Iw again? Is his initial w = 0?

 

[lightgrav]

Before Bob started spinning, wasn't his angular speed = 0?
Before Bob started spinning, was the platform rotating?

The full equation about angular momentum is:
L_initial + Torque_external * duration = L_final .

This should look like Newton's 2nd Law , or linear momentum & impulse

L is always the Sum of Iw terms , if there is more than one object.
(L can also be written as L = r x p , with the vector cross product)

 

[FlipStyle1308]

Hmm... so Ia = L final?

 

[lightgrav]

Ia ?? do you mean I alpha ? No. (change in velocity is not an acceleration)

L_initial = zero , because nothing was moving beforehand.
So the SUM of WHAT "Iw terms" should you set equal to zero?

 

[FlipStyle1308]

0 = Iw + Iw = (2)(70) + (198)w ? So I get a negative w?

 

[lightgrav]

If the Monkey spins clockwise (viewed from above, say)
the platform and he will turn slowly counterclockwise ... Yes!

(To spin clockwise, he pushed on the platform with counter-clockwise Forces)

 

[FlipStyle1308]

Okay, but is my equation correct?

 

[lightgrav]

yes, it appears that the units cancel.

 

[FlipStyle1308]

Thanks, what about the second part of the question:

How long does it take for Bob to get his snack?

 

[Hootenanny]

So I get a negative w?

Yes, because the platform will rotate around the central axis is the opposite direction to Bob's.

~H

 

[Doc Al]

Thanks, what about the second part of the question:

How long does it take for Bob to get his snack?

You've presumably found \omega, now use it. How many radians must he traverse to reach the banana?

 

[FlipStyle1308]

Radians = pi, right?

 

[Hootenanny]

Radians = pi, right?

What do you mean by that?

~H

 

[FlipStyle1308]

He travels a total distance of pi, which is the distance of 1/2 a circle.

 

[Hootenanny]

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

 

[FlipStyle1308]

What other info do I need? Is it w initial and w final? Is the w I found the initial or final?

 

[Hootenanny]

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

~H

 

[FlipStyle1308]

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?

 

[Hootenanny]

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

 

[FlipStyle1308]

Um...so pi = [(0.707 + pi/t)/2]t ?

 

[Hootenanny]

Um...so pi = [(0.707 + pi/t)/2]t ?

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

v = \frac{ds}{dt} \Rightarrow \omega = \frac{d\theta}{dt}

dt = \frac{\theta}{\omega}

Where \omega is the angle of travel in radians.

Can you go from here?

~H

 

[FlipStyle1308]

So w=d/t, t=d/w=pi/0.707?

 

[Hootenanny]

So w=d/t, t=d/w=pi/0.707?

That's what I'd do

~H

 

[FlipStyle1308]

Thank you!