# Homework Help: Particle motion when wrapped around drum; elastic string

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1. Oct 2, 2016

### Dazed&Confused

1. The problem statement, all variables and given/known data
A uniform cylindrical drum of mass $M$ and radius $a$ is free to rotate about its axis, which i is horizontal. An elastic cable of negligible mass and length $l$ is wrapped around the drum and carries on its free end a mass $m$. The cable has elastic potential energy $\tfrac12 kx^2$ where $x$ is the extension of the cable. The cable is initially fully wound up. Show that the motion of the mass $m$ is a uniform acceleration the same as it would be if the cable were inelastic, but it is superimposed with an oscillation of angular frequency given by $\omega^2 = k(M +2m)/Mm$. Find the amplitude of the oscillation if the system is released from rest with the cable unextended.

2. Relevant equations

3. The attempt at a solution

The Lagrangian is
$$L = \tfrac14 Ma^2 \dot{\theta}^2 + \tfrac{m}{2} (a\dot{\theta} +\dot{x})^2 + mg(a\theta +x) -\tfrac12 k x^2$$
where $\theta$ is the angle rotated from its rest position such that a positive angle results in the mass moving downwards. From this we get the equations
$$m(a \ddot{\theta} + \ddot{x} ) =mg -kx$$
and
$$Ma \ddot{\theta} +2m(a \ddot{\theta} +\ddot{x}) = 2mg$$

We can eliminate $\ddot{\theta}$ to get
$$\ddot{x} = -\frac{k(M+2m)}{mM}x + g$$
We can eliminate $g$ by intorducing a new variable but the key point is the angular frequency is the one given in the question. The amplitude then becomes $Mmg/k(M+2m)$, whereas the one given in the book is $M^2mg/k(M+2m)^2$. Solving for $\ddot{\theta}$ I get
$$a \ddot{\theta} = \frac{2k}{M}x$$
which isn't the uniform acceleration they asked for. The uniform acceleration I think should be $2mg/(M+2m)$.

Last edited: Oct 2, 2016
2. Oct 2, 2016

### haruspex

I think you have some signs wrong in your Lagrangian.

I really do not like this question. The expression for elastic PE is completely unrealistic here. It is only correct for uniformly stretched elastic. If k is the elastic constant for the whole string, then for the unwound piece of length y it is kl/y. If the extension is x then the PE is klx2/(2y).

3. Oct 2, 2016

### Dazed&Confused

I've looked at it a bit and I'm not sure what the sign error would be.

4. Oct 2, 2016

### haruspex

Why do you think $a\ddot\theta$ would give the underlying constant acceleration? Can you think of another way to find it?

5. Oct 2, 2016

### Dazed&Confused

I did combine two questions here. The previous question was without the string being inelastic and the constant acceleration term was $a \ddot{\theta}$. I guess there is no other acceleration with the coordinates I have chosen.

6. Oct 2, 2016

### Dazed&Confused

If I find the acceleration $a \ddot{\theta}$ perhaps then combining the two it will be obvious.

7. Oct 2, 2016

### haruspex

You know that x is a sin function of t, plus a constant, no? Can you not simply discard the sine function and solve for $\ddot\theta$?

8. Oct 2, 2016

### Dazed&Confused

I'm not sure what you mean, but I found the answer they were looking for. From $a \ddot{\theta} = 2kx/M$ I found $a \theta$ and then added it to $x$ from which we get a term corresponding to a constant acceleration of $2mg/(M+2m)$ and an oscillation with the same frequency but with amplitude as given in the book.

9. Oct 2, 2016

### haruspex

Ah, of course, it's the amplitude of the aθ+x oscillation you want, which is not the same as that of x.
Well done. Sorry I wasn't more helpful.