Here is a problem I spent my sunday trying to solve. I made progress, but still something is wrong.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A drum of massMand radius_{A}arotates freely with initial angular velocity [tex]\omega_{A}(0).[/tex] A second drum with massMand radius_{B}b>ais mounted on the same axis and is at rest, although it is free to rotate. A thin layer of sand with massMis distributed on the inner surface of the smaller drum. At_{S}t=0small perforations in the inner drum are opened. The sand starts to fly out at a constant rate [tex]\lambda[/tex] and sticks to the outer drum. Find the subsequent angular velocities of the two drums [tex]\omega_{A}[/tex] and [tex]\omega_{B}.[/tex] Ignore the transit time of the sand.

Ans. clue. If [tex]\lambda t = M_{B}[/tex] andb=2athen [tex]\omega_{B}=\omega_{A}(0)/8[/tex]

2. Relevant equations

[tex]\frac{dL}{dt}=0[/tex]

[tex]L=mr^{2}\omega[/tex]

3. The attempt at a solution

I begin by looking at the angular momentum of the inner drum: [tex]L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}\omega_{A}(t)[/tex] and at a later time [tex]L_{A}(t+\Delta t)=(M_{A}+M_{S}-\lambda (t+\Delta t))a^{2}\omega_{A}(t+\Delta t)+\lambda \Delta t a^{2}\omega_{A}(t).[/tex] Taking the limit and dividing bydtgives [tex]\frac{dL_{A}}{dt}=(M_{A}+M{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}-\omega_{A}\lambda a^{2}=0.[/tex] This is a differential equation which can be solved by switching a little to give

[tex]\omega_{A}=\omega_{A}(0)\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}[/tex]

I think this might be right.

However, the second part I am not so sure of. First of all some energy is lost when the sand falls in partly perpendicular to the surface of the outer drum. One can show that a fractiona/bof the momentum survives and the rest becomes heat. Therefore the angular momentum of the outer drum at timetwill be [tex]L_{B}(t)=(\lambda t + M_{B})b^{2}\omega_{B}(t)+b\frac{a}{b}a\omega_{A} \lambda \Delta t,[/tex] where the second term is the incoming sand losing some mechanical energy. At a later time [tex]L_{B}(t)=(\lambda (t+\Delta t) + M_{B})b^{2}\omega_{B}(t+\Delta t).[/tex] Taking the limit and dividing bydtgives [tex]\frac{dL_{B}}{dt}=(\lambda t+M_{B})b^{2}\frac{d\omega_{B}}{dt}+\lambda(\omega_{B}b^{2}-\omega_{A}(0)a^{2}\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t})=0[/tex] This is a linear differential equation which can be solved to give [tex]\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}}\ln \lgroup \frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}\rgroup \frac{M_{A}+M_{S}}{M_{B}+\lambda t}[/tex]

This is not in accordance with the clue, nor with my answer sheet that says [tex]\omega_{B}=\omega_{A}(0)\frac{a^{2}(M_{A}+M_{S}-\lambda t)}{b^{2}(M_{B}+\lambda t)}[/tex]

Please tell me if something is unclear.

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# Conserved angular momentum: finding angular velocities of drums as a function of time

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