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Dazed&Confused
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Homework Statement
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.
Homework Equations
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].
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