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Conserved angular momentum: finding angular velocities of drums as a function of time 
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#1
Jun1911, 02:23 PM

P: 97

Here is a problem I spent my sunday trying to solve. I made progress, but still something is wrong.
1. The problem statement, all variables and given/known data A drum of mass M_{A} and radius a rotates freely with initial angular velocity [tex]\omega_{A}(0).[/tex] A second drum with mass M_{B} and radius 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_{S} is distributed on the inner surface of the smaller drum. At 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. 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 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. 


#2
Jun2011, 12:26 AM

P: 374

Isn't your first differential equation dWa/Wa=lambda*dt/(Ma+Mslambda*t)?



#3
Jun2011, 02:40 PM

P: 97

However I think that the angular momentum of the inner drum might be constant. This is because now looking closer at my equations L_{A}(t) and L_{A}(t+delta*t) I get a sign difference. [tex]L_{A}(t+\Delta t)L_{A}(t)=(M_{A}+M_{S}\lambda t)a^{2}(\omega_{A}(t+\Delta t)\omega_{A}(t))+\lambda a^{2}\Delta t(\omega_{A}(t)\omega_{A}(t+\Delta t))[/tex] Now the second term in this equation becomes zero when taking the limit, so [tex](M_{A}+M_{S}\lambda t)a^{2}\frac{d\omega_{A}}{dt}=0[/tex] with the only solution that the angular momentum is constant. Now putting this into the differential eguation for the second drum [tex]\frac{d\omega_{B}}{dt}+\frac{\lambda}{\lambda t+M_{B}}\omega_{B}=\omega_{A}(0)\frac{a^{2}}{b^{2}}\frac{\lambda}{\lamb da t+M_{B}},[/tex] with the solution [tex]\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}} \frac{\lambda t}{\lambda t+M_{B}}[/tex] This is correct according to the answer clue but not according to my answer sheet. Do you think this might be right? Will a drum losing mass have a constant angular momentum? 


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