Two rotating coaxial drums and sand transfering between them (Kleppner)

In summary: Yep, thanks!In summary, the solution is simple by noting that the total angular momentum of the system is constant. Though I overlooked this, it is still a valid solution.
  • #1
yucheng
231
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!
 
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  • #2
yucheng said:
.
$$L_a(t +\Delta t)= (M_a- \Delta m) a^2 \omega_a(t + \Delta t) + \Delta m a^2 \omega_a(t + \Delta t) = (M_a) a^2 \omega_a(t + \Delta t)$$Thanks in advance!
1. Why would ##\omega_a## change?
2. Why after subtracting ##\Delta ma^2 \omega_a(t + \Delta t)## do you add it on again?
You don't need the ##\Delta t## in there; that would be a second order small quantity.
I assume ##\Delta m= \lambda\Delta t##.
 
  • #3
@haruspex Oops, ##\omega_a(t+\Delta t)## is supposed to be a function of time.

for 2. I was thinking, angular momentum does not change, so... I add that in?

Or, does this angular momentum analysis not work, because I should be considering drum a and drum b at once, not separately?
 
  • #4
yucheng said:
for 2. I was thinking, angular momentum does not change, so... I add that in?
It's losing mass, but its angular momentum doesn't change? Is it going faster?!
 
  • #5
haruspex said:
It's losing mass, but its angular momentum doesn't change? Is it going faster?!
Yes it's going faster, Oh, a contradiction!
 
  • #6
yucheng said:
Oh, a contradiction!
No contradiction, it just loses angular momentum.
yucheng said:
I should be considering drum a and drum b at once, not separately?
Yes, the angular momentum of the combined system is conserved.
 
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  • #7
haruspex said:
No contradiction, it just loses angular momentum.
I mean, if I were to assume my original equation holds, then I can derive a contradiction? Since in my original equation I assumes that angular momentum does not change. While in fact, my conclusion that angular velocity does not change, says it does?
 
  • #8
yucheng said:
I mean, if I were to assume my original equation holds, then I can derive a contradiction? Since in my original equation I assumes that angular momentum does not change. While in fact, my conclusion that angular velocity does not change, says it does?
Ok.
Did you see my edit to post #6?
 
  • #9
Yep, thanks!
haruspex said:
Ok.
Did you see my edit to post #6?
 

What is the purpose of the two rotating coaxial drums in the Kleppner experiment?

The two rotating coaxial drums in the Kleppner experiment are used to demonstrate the transfer of sand between two rotating cylinders, which is a common phenomenon in many industrial processes. This experiment allows scientists to study the dynamics of granular materials and their behavior under different conditions.

How does the sand transfer between the two rotating drums in the Kleppner experiment?

The sand transfer in the Kleppner experiment is driven by the rotation of the drums. As the inner drum rotates, it creates a centrifugal force that pushes the sand towards the outer drum. The sand then falls into the gap between the two drums and is carried by the rotation of the outer drum to the top, where it falls back into the inner drum. This continuous transfer of sand between the two drums is what creates the mesmerizing sand patterns observed in the experiment.

What factors affect the sand transfer in the Kleppner experiment?

There are several factors that can affect the sand transfer in the Kleppner experiment, including the speed and direction of rotation of the drums, the size and shape of the sand particles, and the gap between the two drums. These factors can impact the dynamics of the sand and result in different patterns and behaviors.

What can we learn from studying the sand transfer in the Kleppner experiment?

The Kleppner experiment allows scientists to study the behavior of granular materials, which are commonly found in nature and industrial processes. By understanding the dynamics of sand transfer between the two rotating drums, we can gain insights into the behavior of other granular materials, such as powders, grains, and rocks. This knowledge can be applied to various fields, including geology, engineering, and materials science.

Are there any real-world applications of the Kleppner experiment?

Yes, the principles demonstrated in the Kleppner experiment have real-world applications in industries such as mining, agriculture, and pharmaceuticals. For example, the transfer of granular materials between rotating drums is commonly used in conveyor systems for transporting grains and powders. By studying this phenomenon in the Kleppner experiment, scientists can improve the efficiency and design of these industrial processes.

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