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**1. Homework Statement**

A drum of mass

*M*and radius

_{A}*a*rotates freely with initial angular velocity [tex]\omega_{A}(0).[/tex] A second drum with mass

*M*and radius

_{B}*b>a*is mounted on the same axis and is at rest, although it is free to rotate. A thin layer of sand with mass

*M*is distributed on the inner surface of the smaller drum. At

_{S}*t=0*small 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] and

*b=2a*then [tex]\omega_{B}=\omega_{A}(0)/8[/tex]

**2. Homework 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 by

*dt*gives [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 fraction

*a/b*of the momentum survives and the rest becomes heat. Therefore the angular momentum of the outer drum at time

*t*will 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 by

*dt*gives [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.