- #1
yucheng
- 232
- 57
- Homework Statement
- A drum of mass ##M_A## and radius ##a## rotates freely with initial angular speed ##\omega _0##. A second drum of radius ##b>a## and mass ##M_B## is mounted on the same axis, although it is free to rotate. A thin layer of sand ##M_s## is evenly distributed on the inner surface of the smaller drum (drum ##A##). At ##t=0## small perforations in the inner drum are opened, and sand starts to fly out at a rate ##\dfrac{dM}{dt}=\lambda## and sticks to the outer drum. Find subsequent angular velocities of the two drums. Ignore the transit time of the sand
Kleppner, Problem 7.2
- Relevant Equations
- N/A
The solution is simple by noting that the total angular momentum of the system is constant. (Though I overlooked this)
Instead, I went ahead analyzing the individual angular momentum of both drums.
Let ##L_a## and ##L_b## be the angular momentum respectively. ##M_a##, ##M_b## be the instantaneous mass (including sand), ##M_{0a}##, ##M_{0b}## be initial mass.
Then $$L_a(t) = M_a a^2 \omega_a(t)$$ and $$L_a(t +\Delta t)= (M_a- \Delta m) a^2 \omega_a(t + \Delta t) + \Delta m a^2 \omega_a(t + \Delta t) = (M_a) a^2 \omega_a(t + \Delta t)$$
$$\Delta L_a = M_a a^2 (\omega_a(t + \Delta t) - \omega_a(t))$$
$$\frac{\Delta L_a}{\Delta t} = M_a a^2 \frac{\omega_a(t + \Delta t) - \omega_a(t)}{\Delta t}$$
Taking the limit, $$\frac{d L_a}{dt} = M_a a^2 \frac{d \omega_a}{dt}$$
But since there is no net torque, $$ \frac{d \omega_a}{dt} = \frac{d L_a}{dt} = 0$$, thus ##\omega_a## is constant.
This fits our prediction, but is this proof correct?
Next, we do the same for drum B, we get
$$L_b(t) = M_b b^2 \omega_b(t) + \Delta m a^2 \omega_a$$
$$L_b(t +\Delta t)= (M_b + \Delta m) b^2 (\omega_b(t) + \delta \omega_b)$$
Can I write ##\omega_b## instead of ##\omega_b(t)##?
Ignoring second order terms,
$$\Delta L_b = b^2 M_b \Delta \omega_b + \Delta m (b^2 \omega_b - a^2 \omega_a)$$
$$\frac{\Delta L_b}{\Delta t}= b^2 M_b \frac{\Delta \omega_b}{\Delta t} + \frac{\Delta m}{\Delta t} (b^2 \omega_b - a^2 \omega_a)$$
Taking the limits,
$$\frac{d L_b}{d t}= b^2 M_b \frac{d \omega_b}{d t} + \frac{d m}{d t} (b^2 \omega_b - a^2 \omega_a)$$
Since $$L_b=I \omega_b= M_b b^2 \omega_b$$, I thought of solving the differential equation:
$$d L_b= (b^2 M_b \frac{d \omega_b}{d t} + \frac{d m}{d t} (b^2 \omega_b - a^2 \omega_a))dt$$
But ##M_b=M_0b + \lambda b t##
$$d L_b= (b^2 (M_b=M_{0b} + \lambda b t)\frac{d \omega_b}{d t} + \lambda (b^2 \omega_b - a^2 \omega_a))dt$$
How should I go about integrating this equation?
Thanks in advance!
Instead, I went ahead analyzing the individual angular momentum of both drums.
Let ##L_a## and ##L_b## be the angular momentum respectively. ##M_a##, ##M_b## be the instantaneous mass (including sand), ##M_{0a}##, ##M_{0b}## be initial mass.
Then $$L_a(t) = M_a a^2 \omega_a(t)$$ and $$L_a(t +\Delta t)= (M_a- \Delta m) a^2 \omega_a(t + \Delta t) + \Delta m a^2 \omega_a(t + \Delta t) = (M_a) a^2 \omega_a(t + \Delta t)$$
$$\Delta L_a = M_a a^2 (\omega_a(t + \Delta t) - \omega_a(t))$$
$$\frac{\Delta L_a}{\Delta t} = M_a a^2 \frac{\omega_a(t + \Delta t) - \omega_a(t)}{\Delta t}$$
Taking the limit, $$\frac{d L_a}{dt} = M_a a^2 \frac{d \omega_a}{dt}$$
But since there is no net torque, $$ \frac{d \omega_a}{dt} = \frac{d L_a}{dt} = 0$$, thus ##\omega_a## is constant.
This fits our prediction, but is this proof correct?
Next, we do the same for drum B, we get
$$L_b(t) = M_b b^2 \omega_b(t) + \Delta m a^2 \omega_a$$
$$L_b(t +\Delta t)= (M_b + \Delta m) b^2 (\omega_b(t) + \delta \omega_b)$$
Can I write ##\omega_b## instead of ##\omega_b(t)##?
Ignoring second order terms,
$$\Delta L_b = b^2 M_b \Delta \omega_b + \Delta m (b^2 \omega_b - a^2 \omega_a)$$
$$\frac{\Delta L_b}{\Delta t}= b^2 M_b \frac{\Delta \omega_b}{\Delta t} + \frac{\Delta m}{\Delta t} (b^2 \omega_b - a^2 \omega_a)$$
Taking the limits,
$$\frac{d L_b}{d t}= b^2 M_b \frac{d \omega_b}{d t} + \frac{d m}{d t} (b^2 \omega_b - a^2 \omega_a)$$
Since $$L_b=I \omega_b= M_b b^2 \omega_b$$, I thought of solving the differential equation:
$$d L_b= (b^2 M_b \frac{d \omega_b}{d t} + \frac{d m}{d t} (b^2 \omega_b - a^2 \omega_a))dt$$
But ##M_b=M_0b + \lambda b t##
$$d L_b= (b^2 (M_b=M_{0b} + \lambda b t)\frac{d \omega_b}{d t} + \lambda (b^2 \omega_b - a^2 \omega_a))dt$$
How should I go about integrating this equation?
Thanks in advance!