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I came across a simpler version of this question where the man was stationary, and pulling on a string which was wrapped around a cylinder. We assume that there is no slipping anywhere.

enter image description here

Then we know for pure rolling, the top most point of the cylinder would be moving with double the velocity of the Velocity of centre of mass of the cylinder (Since the top most point would posess both rotational+translational speeds). That is, if

Velocity of centre of mass of cylinder $=V$
Then, Velocity of top most point $=2V$

Hence the string would move twice as fast. Therefore if the cylinder travelled a distance $l$, the length of string that would pass the man's hands would be $2l$.


If we took this problem a bit further and say, the man also starts moving away/towards the cylinder while pulling the string, what would happen now?

enter image description here

I thought maybe if we looked from the frame of the man, the cylinder would seem to be coming with a speed: $2V$ , Hence the top most point would seem to be coming with a velocity $4V$ hence if the cylinder travelled a length $l$, the length of string that would pass the man's hands would be $4l$.

But this still isn't very clear to me, and I feel this was just a vague guess of mine. I would appreciate if someone could make things a bit more clear for me and how can I approach this problem?

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If the man also walks towards the cylinder with speed $V$, the velocity of the centre of the cylinder relative to him is $V--V=2V$, as you have asserted.

However, the velocity of the top of the cylinder relative to the man is $2V--V=3V$ not $4V$.

So if the cylinder travelled a distance of $l$, a length of $3l$ of rope would have passed through the man's hands.

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