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Particle motion when wrapped around drum; elastic string

  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform cylindrical drum of mass [itex] M [/itex] and radius [itex] a [/itex] is free to rotate about its axis, which i is horizontal. An elastic cable of negligible mass and length [itex]l[/itex] is wrapped around the drum and carries on its free end a mass [itex]m[/itex]. The cable has elastic potential energy [itex]\tfrac12 kx^2 [/itex] where [itex]x[/itex] is the extension of the cable. The cable is initially fully wound up. Show that the motion of the mass [itex]m[/itex] 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 [itex] \omega^2 = k(M +2m)/Mm [/itex]. 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
    [tex]
    L = \tfrac14 Ma^2 \dot{\theta}^2 + \tfrac{m}{2} (a\dot{\theta} +\dot{x})^2 + mg(a\theta +x) -\tfrac12 k x^2
    [/tex]
    where [itex]\theta[/itex] 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
    [tex]
    m(a \ddot{\theta} + \ddot{x} ) =mg -kx
    [/tex]
    and
    [tex]
    Ma \ddot{\theta} +2m(a \ddot{\theta} +\ddot{x}) = 2mg
    [/tex]

    We can eliminate [itex]\ddot{\theta}[/itex] to get
    [tex]
    \ddot{x} = -\frac{k(M+2m)}{mM}x + g
    [/tex]
    We can eliminate [itex]g[/itex] by intorducing a new variable but the key point is the angular frequency is the one given in the question. The amplitude then becomes [itex]Mmg/k(M+2m)[/itex], whereas the one given in the book is [itex]M^2mg/k(M+2m)^2[/itex]. Solving for [itex]\ddot{\theta}[/itex] I get
    [tex]
    a \ddot{\theta} = \frac{2k}{M}x
    [/tex]
    which isn't the uniform acceleration they asked for. The uniform acceleration I think should be [itex] 2mg/(M+2m) [/itex].
     
    Last edited: Oct 2, 2016
  2. jcsd
  3. Oct 2, 2016 #2

    haruspex

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    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).
     
  4. Oct 2, 2016 #3
    I've looked at it a bit and I'm not sure what the sign error would be.
     
  5. Oct 2, 2016 #4

    haruspex

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    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?
     
  6. Oct 2, 2016 #5
    I did combine two questions here. The previous question was without the string being inelastic and the constant acceleration term was [itex] a \ddot{\theta}[/itex]. I guess there is no other acceleration with the coordinates I have chosen.
     
  7. Oct 2, 2016 #6
    If I find the acceleration [itex]a \ddot{\theta}[/itex] perhaps then combining the two it will be obvious.
     
  8. Oct 2, 2016 #7

    haruspex

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    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##?
     
  9. Oct 2, 2016 #8
    I'm not sure what you mean, but I found the answer they were looking for. From [itex] a \ddot{\theta} = 2kx/M [/itex] I found [itex] a \theta[/itex] and then added it to [itex]x[/itex] from which we get a term corresponding to a constant acceleration of [itex]2mg/(M+2m)[/itex] and an oscillation with the same frequency but with amplitude as given in the book.
     
  10. Oct 2, 2016 #9

    haruspex

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    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.
     
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