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

 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 MA and radius a rotates freely with initial angular velocity $$\omega_{A}(0).$$ A second drum with mass MB 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 MS 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 $$\lambda$$ and sticks to the outer drum. Find the subsequent angular velocities of the two drums $$\omega_{A}$$ and $$\omega_{B}.$$ Ignore the transit time of the sand. Ans. clue. If $$\lambda t = M_{B}$$ and b=2a then $$\omega_{B}=\omega_{A}(0)/8$$ 2. Relevant equations $$\frac{dL}{dt}=0$$ $$L=mr^{2}\omega$$ 3. The attempt at a solution I begin by looking at the angular momentum of the inner drum: $$L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}\omega_{A}(t)$$ and at a later time $$L_{A}(t+\Delta t)=(M_{A}+M_{S}-\lambda (t+\Delta t))a^{2}\omega_{A}(t+\Delta t)+\lambda \Delta t a^{2}\omega_{A}(t).$$ Taking the limit and dividing by dt gives $$\frac{dL_{A}}{dt}=(M_{A}+M{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}-\omega_{A}\lambda a^{2}=0.$$ This is a differential equation which can be solved by switching a little to give $$\omega_{A}=\omega_{A}(0)\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}$$ 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 $$L_{B}(t)=(\lambda t + M_{B})b^{2}\omega_{B}(t)+b\frac{a}{b}a\omega_{A} \lambda \Delta t,$$ where the second term is the incoming sand losing some mechanical energy. At a later time $$L_{B}(t)=(\lambda (t+\Delta t) + M_{B})b^{2}\omega_{B}(t+\Delta t).$$ Taking the limit and dividing by dt gives $$\frac{dL_{B}}{dt}=(\lambda t+M_{B})b^{2}\frac{d\omega_{B}}{dt}+\lambda(\omega_{B}b^{2}-\omega_{A}(0)a^{2}\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t})=0$$ This is a linear differential equation which can be solved to give $$\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}}\ln \lgroup \frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}\rgroup \frac{M_{A}+M_{S}}{M_{B}+\lambda t}$$ This is not in accordance with the clue, nor with my answer sheet that says $$\omega_{B}=\omega_{A}(0)\frac{a^{2}(M_{A}+M_{S}-\lambda t)}{b^{2}(M_{B}+\lambda t)}$$ Please tell me if something is unclear.
Yes it is $$\frac{d\omega_{A}}{\omega_{A}}=\lambda \frac{dt}{M_{A}+M_{S}-\lambda t}$$ This leads to $$\ln \omega_{A}=-\ln (M_{A}+M_{S}-\lambda t)+C$$ which leads to my equation stated in original post.
However I think that the angular momentum of the inner drum might be constant. This is because now looking closer at my equations LA(t) and LA(t+delta*t) I get a sign difference. $$L_{A}(t+\Delta t)-L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}(\omega_{A}(t+\Delta t)-\omega_{A}(t))+\lambda a^{2}\Delta t(\omega_{A}(t)-\omega_{A}(t+\Delta t))$$ Now the second term in this equation becomes zero when taking the limit, so $$(M_{A}+M_{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}=0$$ with the only solution that the angular momentum is constant.
Now putting this into the differential eguation for the second drum $$\frac{d\omega_{B}}{dt}+\frac{\lambda}{\lambda t+M_{B}}\omega_{B}=\omega_{A}(0)\frac{a^{2}}{b^{2}}\frac{\lambda}{\lamb da t+M_{B}},$$ with the solution $$\omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}} \frac{\lambda t}{\lambda t+M_{B}}$$ 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?