Moment of Inertia of Massless Pulley w/String & Mass m Attached

[James1234567]

What is the moment of inertia of massless (ideal) pulley that has a string attached to its center and a mass m attached to that string?

If this pulley were hanging from a massless rope that was attached on one side to the cieling and the other side had a tension of mg, what would be the acceleration (upwards) of the mass attached to the pulley?

Thanks for any insight.

 

[BvU]

Hello James,

You are asking for some views in response to your questions; it doesn't look like homework, so I'll assume you are being introduced to the matter and have some things you want to clear up.

The moment of inertia about the central axis of a massless pulley is the sum of all ##\ r^2\; \Delta m\ ## contributions of the little chunks ##\ \Delta m\ ## from which it is built up. Massless means this gives zero. If you want to add the string, for that ##\ r \ ## is zero, so it doesn't contribute either.

And the mass attached to the center of the pulley stays in place if the thing starts to turn like mad ...

But maybe I have the wrong picture in mind, so perhaps you can put me right by providing a better one ?

 

[BvU]

Well, that was a double-post. Let me produce a more benevolent interpretation:

and now we ask: what is the tension in the left piece of rope ?

Still sure this isn't homework ?

 

[James1234567]

Well, that was a double-post. Let me produce a more benevolent interpretation: View attachment 104965 and now we ask: what is the tension in the left piece of rope ? Still sure this isn't homework ?

Well, that was a double-post. Let me produce a more benevolent interpretation: View attachment 104965 and now we ask: what is the tension in the left piece of rope ? Still sure this isn't homework ?

 

[James1234567]

Well, that was a double-post. Let me produce a more benevolent interpretation: View attachment 104965 and now we ask: what is the tension in the left piece of rope ? Still sure this isn't homework ?

Well, that was a double-post. Let me produce a more benevolent interpretation: View attachment 104965 and now we ask: what is the tension in the left piece of rope ? Still sure this isn't homework ?

 

[James1234567]

BvU, you've nearly got the picture perfect. If you remove the circle (or pulley) at the bottom and attach the string directly to the mass m (the rectangle), it will be perfect.

So here come some questions:
So based on what you wrote and what I've learned, the pulley has no moment of inertia. That would mean that the tension throughout the rope slung around the pulley is unoform. Since the tension on the right is mg, then the tension on the left is a part of that mg. And then because the force downward due to the mass is mg, the mass would not accelerate, correct?

At least that's what my logic is, but I believe it has to accelerate in this case, so any help would be appreciated.

Sorry BvU, I should have been more direct in my first post: this is a practice problem a teacher gave me years ago. I just pulled out some old high school physics stuff I had saved and was looking through it when I found some problems I didn't have solutions for anymore. I'm just looking to figure it out and refresh my physics memory a bit while I clean out some old stuff.

 

[BvU]

Fair enough. (the circle in the drawing was the ring at the top of the old-fashioned weight

)

Since the tension on the right is mg, then the tension on the left is a part of that mg

Try the following experiment: take a piece of rope and pull one side with a different force than on the other side. Or the other way around: what would be the acceleration of the rope if it were massless and the forces were different ?
Draping a rope over a frictionless pulley doesn't change the situation, so:

After this intermezzo: what is the tension on the left side of the rope ?

Subsequently you draw a free-body diagram of the pulley. It's massless, so to have a finite acceleration the forces must balance.

Finally you draw a free-body diagram of the weight and calculate its acceleration.

Best of wishes with resuming your endeavours in this highly exciting field of science !

 

[James1234567]

So what you are saying is that the tension is uniform throughout the rope, right? If I pull both ends of the rope, I end up with a rope with uniform tension. So the tension on the left is 1/2 mg and the tension on the right is 1/2 mg, meaning the mass doesn't accelerate?

This doesn't seem right to me. I apologize if I'm frustrating you with my lack of memory on the subject, but I haven't looked at this stuff in quite a long time.

 

[BvU]

The original exercise text had

on one side to the ceiling and the other side had a tension of mg

So mg on both sides.

 

[BvU]

The mass of the pulley is given: zero ! We were getting there at our own pace...

 

[CWatters]

Have deleted my posts. Sorry for the interruption!

 

[James1234567]

Wouldn't that mean the mass is accelerating upwards at a=g? I would have expected that the answer would have to do with the rotation of the pulley on the rope.

Sorry, but can you explain the logic behind the fact that the left side has a tension of mg?

 

[BvU]

The tension in the rope of mg on the righthand side is equal to the tension on the lefthand side.
All the pulley does with the rope is pass on this tension over an angle of 180 degrees.
The two tensions combine to an upward force of 2mg.
A free body diagram of the pulley makes clear the 2mg upward must be offset by 2mg downward that can only come from the rope from which the weight hangs.
So the tension in that rope must be 2mg.
A free body diagram of the weight then reveals that: yes, the mass is accelerating upward at a=g.

can you explain the logic behind the fact that the left side has a tension of mg?

It is not possible to exert a different tension on one end of a rope than on the other (ideal, massless,...). Wrapped around a frictionless (or massless) pulley or not. (You did the experiment ?)

 

[James1234567]

Haha, yes I did the experiment. :P

I think I've got it now. I had to re-read this whole conversation a few times, but I think I've got it. I can't thank you enough BvU, you've been so extremely helpful and patient. I think I had just spent too much time looking over old physics stuff and convinced myself that every problem needed to be more complicated than that. (i.e. involve rotation, moment of inertia, etc.) hopefully we'll meet again sometime on the internet - maybe then I can pay you back with some help in some way or another. Anyway, the point is: Thank you!

 

[BvU]

You're welcome. My pleasure.