# Small cylinder inside a larger one, rolling without slipping

1. Feb 8, 2014

### peripatein

Hi,
Please note: this is not a homework assignment. It is taken from an exam and I'd appreciae some clarifications.
In the setup delineated in the attachment, a small cylinder of mass M and radius R rolls without slipping inside a larger cylinder of radius 10R. It is stated that the larger cylinder doesn't move.
Now, if linear momentum is conserved along the x axis, due to the absence of external forces along that axis, and is equal to zero at the beginning, how come v1 would not be equal to zero as well? I know how to find v1 using conservation of energy and know it is not equal to zero, but do not understand why conservation of linear momentum would not entail v1=0. I'd appreciate an explanation.

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2. Feb 8, 2014

### voko

The fact that the outer cylinder does not move means there is an external force that keeps it in place.

3. Feb 8, 2014

### BvU

Momentum is a vector. I am surprised why your v1=0 expectation isn't accompanied by an expectation that v0=0 and remains 0 as well ? After all, the linear momentum is conserved, or perhaps it is not ? (*)

This was a rhetorical question. Of course you don't expect that v0 doesn't change, because there are forces at work. Even is v0=0 at t=t0, the earth's gravitational field does work on the small cylinder. Something is holding the big cylinder in place, so there's no work done by that something ($d\vec s =\vec 0$).

Big cylinder itself doesn't do any work either: all it can do is exercise a force to keep small cylinder on a circular path and that means the acceleration (ergo force) is always perpendicular to the motion: $\vec F \cdot d \vec s = 0$. But the non-working force does change the direction of the motion. So we have no conservation of linear momentum anywhere: gravity does work (you'd have to establish a coordinate system to give it a sign) one half of the period, gets it back the other half.

So much for the simple picture. "Aha!", you say: the friction is not perpendicular to the motion, so it does do work! Not so: near the contact point motion is perpendicular to the inner wall of the big cylinder (trajectory is a cycloid). Phew...

It does execute a torque and it does cause angular acceleration but I'm confident that the work to change rotation energy is being done by gravity force alone. So confident that I'm too lazy to work it out. (Does mean that v1 is smaller in the non-slipping case than in the in the no-friction case.)

Slight criticism (on an exam exercise, sic!): As it is drawn in the top figure it is very difficult to imagine how the non-slipping is achieved (well: see *). This can distract the best of candidates.

(*) I can re-read the OP and take it exactly for what it is. In that case the v0=0 is unrealistic: it has to be so big that even at the top of big cylinder the net normal force (pointing down) is big enough to achieve the non-slipping. Looking at it that way makes it a realistic exercise. Story about change in direction holds. Angular acceleration/deacceleration caused by gravity pointing in the same direction all the time makes |v1|>|v0| but you already had that from energy balancing.

I hope I have given some clarification, but I wouldn't be surprised if there were more questions now than you had originally. Good. ( :-) also wouldn't be surprised if I uttered nonsense at some point(s?), so please correct me there!)

Grab a piece of drain pipe and a piece of gas pipe and experiment, is what I say. But then, I'm an experimental physicist, and you can't do that during an exam...

Last edited: Feb 8, 2014