# Pulling a cylinder - find final angular velocity

• RightFresh

## Homework Statement

A massless rope is wrapped several times around a solid cylinder of radius R = 20 cm and
mass M = 20 kg, which is at rest on a horizontal surface.
Someone pulls 1 m of the rope with a constant force of 100 N, setting the cylinder in motion.
Assuming that the rope neither stretches nor slips and that the cylinder rolls without slipping,
what is the final angular velocity of the cylinder and the speed at its surface?
The moment of inertia of a cylinder of mass M and radius R is MR2/2.

## Homework Equations

work=force*distance=Fx
KE_trans=0.5*m*v^2
KE_rot=0.5*I*w^2

## The Attempt at a Solution

I've used conservation of energy to write that Fx=0.5*m*w^2*R^2+0.5*0.5*m*R^2*w^2=0.75*m*R^2*w^2.

Obviously I'd rearrange this for w, but I don't know how to find x. The question says that 1m of rope is pulled off the cylinder, but how do I find the distance that the cylinder travels in this time? I realize this is probably obvious

Hello Fresh, and welcome to PF :)

Does it say in which direction the rope is pulled ? Like: "straight up", for instance ?

For your answer, you'll need something to link angular velocity and speed. The no slip condition helps you out with that.
And the energy equation is spot on. Except...

 It's even better than I first thought (casual reader). Except for one 0.5 I don't get. (casual reader Squared, sorry..)

Work done = Ekinetic, rotation + Ekinetic, translation

## F\;x=0.5\; mR^2\;\omega^2+{\bf0.5} \ 0.5\;m (\omega R)^2=0.75 m \omega R^2 ##

I'm a slow typist. What is this first 0.5 (I tried to get it in bold face) for ?

What I missed is what you mean with x . The exercise asks for v, doesn't it ?

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Hello Fresh, and welcome to PF :)

Does it say in which direction the rope is pulled ? Like: "straight up", for instance ?

For your answer, you'll need something to link angular velocity and speed. The no slip condition helps you out with that.
And the energy equation is spot on.
Thanks for the welcome :)

It's pulled horizontally. And the relation you're talking about is v=Rw, but I'm not sure how to use that to find x

Hello Fresh, and welcome to PF :)

Does it say in which direction the rope is pulled ? Like: "straight up", for instance ?

For your answer, you'll need something to link angular velocity and speed. The no slip condition helps you out with that.
And the energy equation is spot on.

 It's even better than I first thought (casual reader).

Work done = Ekinetic, rotation + Ekinetic, translation

## F\;x=0.5\; mR^2\;\omega^2+{\bf0.5} \ 0.5\;m (\omega R)^2=0.75 m \omega R^2 ##

I'm a slow typist. What is this first 0.5 (I tried to get it in bold face) for ?

What I missed is what you mean with x . The exercise asks for v, doesn't it ?
The first 0.5 comes from rotational ke=0.5*I*w^2, and I=0.5*m*R^2 for a cylinder. The question asks for the final angular velocity and so I set the final energy equal to the work done (torque*angle=F*R*x), where x is the distance the cylinder turns through

Thanks for the welcome :)

It's pulled horizontally. And the relation you're talking about is v=Rw, but I'm not sure how to use that to find x

Oh, boy, now I see: pulling 1 m of rope with 100 N does not mean doing 100 J of work for you ?
That what you are referring to when you say you don't know how to find x ?

The rope is horizontal, 20 cm above the surface and the cylinder rolls towards the someone!

Oh, boy, now I see: pulling 1 m of rope with 100 N does not mean doing 100 J of work for you ?
That what you are referring to when you say you don't know how to find x ?

The rope is horizontal, 20 cm above the surface and the cylinder rolls towards the someone!
I tried that, it gives the wrong answer. The problem is that when pulling 1m of rope, the cylinder will turn through a distance > 1m and this distance is what I call x

You are absolutely correct about the factors 0.5. I should not take any more coffee and go to sleep.
The smiley is for egg on face $$F\;x={\rm Work = E_{kin} + E_{rot} } \;= 0.5\; m(\omega R) ^2\;+\; 0.5\; I\omega^2 = 0.5\; m(\omega R) ^2\;+\;0.5\;0.5mR^2 \omega^2=0.75 m \omega R^2$$

You are absolutely correct about the factors 0.5. I should not take any more coffee and go to sleep.
The smiley is for egg on face $$F\;x={\rm Work = E_{kin} + E_{rot} } \;= 0.5\; m(\omega R) ^2\;+\; 0.5\; I\omega^2 = 0.5\; m(\omega R) ^2\;+\;0.5\;0.5mR^2 \omega^2=0.75 m \omega R^2$$
I'm really confused as to what you are saying, sorry

Sorry to be so slow... bedtime too...

What I'm saying that only now I understand your predicament. Your expression is spot on (first reactions are often the best...).
And x is not immediately obviously 1 meter. It can be > 1m and you wanted to know how much bigger.

Well, experiment can decide ! (about the factor between x and the 1 m). Take something cylindrical and a piece of wire !

In the mean time our team of mathematicians hasn't been doing nothing (they abhor experiments and much rather do it numerically -- or at least calculate :) )

So:

Good thing you have the 'correct' answer. (Sure of that?)

18.26 rad/s would correspond to 200 J of total kinetic energy (trans+rot).

It now comes down to what is meant by
Someone pulls 1 m of the rope
, becasue to me it's ambiguous.

100 J is what you get when you pull 1m of rope: suppose the rope is fed through a hole and the cylinder is 2m further on the other side. There's 25 cm of rope to pull on and the someone pulls until it's 125 cm.

200 J is what you get when you pull 1m of the rope: have a red mark on the rope wherer it's at the top point of the cyliinder. And someone pulls (let's say to the left) until the red mark is 1 m away from the top point of the cylinder.

The someone has stepped back 1m (and to the side as well, if he (or she) is smart :) )

Yuck :(

This guy really isn't helping...sorry. Can anyone else help?

But he's really trying ! It's not all bad, I should hope... :(

Let's ask a good one: @haruspex !

I tried that, it gives the wrong answer. The problem is that when pulling 1m of rope, the cylinder will turn through a distance > 1m and this distance is what I call x

You can do some little experiment...
If you pull off L length of rope from the cylinder (of radius R) it turns θ=L/R angle. The cylinder rolls without slipping. It means that the centre moves with speed Rω, and the distance traveled by the centre is θR.
The work is force times the distance traveled by the point of application. The force is applied at the end of the rope. The cylinder travels θR distance when length L is pulled off. How far is the end of the rope from the initial position?

You can do some little experiment...
If you pull off L length of rope from the cylinder (of radius R) it turns θ=L/R angle. The cylinder rolls without slipping. It means that the centre moves with speed Rω, and the distance traveled by the centre is θR.
The work is force times the distance traveled by the point of application. The force is applied at the end of the rope. The cylinder travels θR distance when length L is pulled off. How far is the end of the rope from the initial position?
2 metres?

2 metres?
Yes! Yay! Thanks :)

Interestingly, this gives me 18.3 not 18.4 rad/s but oh well

• ehild
Will you put in a good word for me, Ehild? A long time I was at the other side of the fence, spending only 100 J.

And Right one, you've got some good assistance here, from someone who can draw fast, too ! :) Will you put in a good word for me, Ehild?
I did not notice your post when I submitted my one. Everything happened so fast :) And you wrote that the work had to be 200 J ...
But an experiment (a real one) is always useful. I have a roll of paper for such rolling problems. :)

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