Particle motion when wrapped around drum; elastic string

[Dazed&Confused]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

The Lagrangian is
<br /> L = \tfrac14 Ma^2 \dot{\theta}^2 + \tfrac{m}{2} (a\dot{\theta} +\dot{x})^2 + mg(a\theta +x) -\tfrac12 k x^2<br />
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
<br /> m(a \ddot{\theta} + \ddot{x} ) =mg -kx<br />
and
<br /> Ma \ddot{\theta} +2m(a \ddot{\theta} +\ddot{x}) = 2mg<br />

We can eliminate \ddot{\theta} to get
<br /> \ddot{x} = -\frac{k(M+2m)}{mM}x + g<br />
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
<br /> a \ddot{\theta} = \frac{2k}{M}x<br />
which isn't the uniform acceleration they asked for. The uniform acceleration I think should be 2mg/(M+2m).

 

[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).

 

[Dazed&Confused]

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).

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

 

[haruspex]

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

Sorry, your signs are correct. I also agree with your amplitude.
Why do you think $a\ddot\theta$ would give the underlying constant acceleration? Can you think of another way to find it?

 

[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.

 

[Dazed&Confused]

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

 

[haruspex]

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

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 $\ddot\theta$?

 

[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.

 

[haruspex]

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.

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.