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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**