Angular momentum, transfer of mass between two drums.

[Slickepot]

Homework Statement

A drum of mass M_A and radius a rotates freely with initial angular velocity ω_A(0). 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 λ and sticks to the outer drum. Find the subsequent angular velocities of the two drums ω_A and ω_B. Ignore the transit time of the sand.
(picture available if needed)

Homework Equations

Angular momentum L=Iω
I for these drums , I=R²Mω

The Attempt at a Solution

At t=0, L(0)=a²(M_A+M_S)ω_A(0).

and at L(t) = a²(M_A+M_S-λt)ω_A+b²(M_B+λt)ω_B

conservation of angular momentum for the system, L(0)=L(t).
and put λt=M_B and b=2a (because of answer hint).

I arrive at,
(M_A+M_S)(ω_A(0)-ω_A)+M_Bω_A=8M_Bω_B

and answer to this hint is ω_B=ω_A(0)/8

which I would get if ω_A(0)-ω_A = 0

Any suggestions what I'm missing?

 

[collinsmark]

Homework Statement

A drum of mass M_A and radius a rotates freely with initial angular velocity ω_A(0). 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 λ and sticks to the outer drum. Find the subsequent angular velocities of the two drums ω_A and ω_B. Ignore the transit time of the sand.
(picture available if needed)

Homework Equations

Angular momentum L=Iω
I for these drums , I=R²Mω

The Attempt at a Solution

At t=0, L(0)=a²(M_A+M_S)ω_A(0).

and at L(t) = a²(M_A+M_S-λt)ω_A+b²(M_B+λt)ω_B

conservation of angular momentum for the system, L(0)=L(t).

Oooh, I like this problem! It's very insightful. By the way, I think you're doing just fine so far.

and put λt=M_B and b=2a (because of answer hint).

I arrive at,
(M_A+M_S)(ω_A(0)-ω_A)+M_Bω_A=8M_Bω_B

and answer to this hint is ω_B=ω_A(0)/8

which I would get if ω_A(0)-ω_A = 0

Any suggestions what I'm missing?

Regarding that last statement, ω_A(0)-ω_A = 0, if correct it means ω_A = ω_A(0). In other words, if the statement is correct, it's implying that drum A never changes its angular velocity, and rotates around at a constant velocity ω_A(0) whether the sand is flying out of it or not.

The next step is to figure out whether that's actually true or not, and if it is true, to prove it (at least prove it to yourself).

You know that the angular momentum of the drum A plus sand combination changes over time, but that might simply be because the combination's mass, thus moment of inertia, is changing, perhaps. But let's step away from angular momentum for a moment and focus on kinematics and Newton's laws of motion, as applied to angular motion.

Ask yourself the following:

1. What is Newton's first law of motion? (Be sure to use the angular motion equivalent.)
2. What, if any, torques are acting on drum A?

 

[Slickepot]

Thanks and right, exactly. I got that far, and then...
When sand leave drum A, it doesn't exert any torque on drum A and drum A doesn't exert and torque on the rest of the system. If no forces act on the drum it will keep on spinning about the same angular velocity. (Newton.1)
Is this enough of an argument?

What threw me off abit is when I thought about the formulas
$$\tau=\frac{dL}{dt}=\frac{d(I\omega)}{dt}=\frac{dI}{dt}+\frac{d\omega}{dt}$$.

And since,
$$\frac{dI}{dt}=\frac{dm}{dt}+\frac{dR^2}{dt}$$ is not zero there would be a torque.

But this would be the sand system, which then exerts this torque on drum B.

This problem sure has a lot of physics in it.

 

[collinsmark]

Thanks and right, exactly. I got that far, and then...
When sand leave drum A, it doesn't exert any torque on drum A and drum A doesn't exert and torque on the rest of the system. If no forces act on the drum it will keep on spinning about the same angular velocity. (Newton.1)
Is this enough of an argument?

What threw me off abit is when I thought about the formulas
$$\tau=\frac{dL}{dt}=\frac{d(I\omega)}{dt}=\frac{dI}{dt}+\frac{d\omega}{dt}$$.

And since,
$$\frac{dI}{dt}=\frac{dm}{dt}+\frac{dR^2}{dt}$$ is not zero there would be a torque.

I think you mean (using the chain rule),

$$\tau=\frac{dL}{dt}= \frac{d(I\omega)}{dt}= \omega \frac{dI}{dt}+I\frac{d\omega}{dt}$$

and

$$\frac{dI}{dt}=R^2 \frac{dm}{dt}+ m\frac{dR^2}{dt}$$

But even with that correction, this has me scratching my head too on how to model the sand mathmatically.

I think we can convince ourselves that the drum A rotates at a constant velocity in a different way.

Consider an ice-skater holding on to two weights, one in each hand, as she goes into a spin. She can change here angular velocity by extending her arms in and out. But now suppose she's spinning at some constant arm length extension (so her angular velocity is constant) and instantaneously releases both weights.

Let's take this thought experiment a step further. Suppose that instead of holding on to each weight with her hands, the weights are attached to a string. The skater and the attached weights are all spinning around together, and the skater is lightly holding onto the center of the string with her finger and thumb (ignore gravity if you wish, or assume that she's spinning around in outer-space or something). Now she carefully cuts the string between her finger and thumb with a pair of scissors and the weights go flying off. I think you can see that it's safe to say that cutting the string in this case is not going to create a torque on the skater, and she will continue spinning at the same constant angular velocity.

But this would be the sand system, which then exerts this torque on drum B.

The sand does in fact create a torque on drum B, because when the sand strikes the inner surface of drum B it has a tangential component to it (which is true as long as the radius of drum A is greater than 0).

[Edit: that is there is a torque on drum B for a > 0, *and* up to some limit for ω_B. Eventually after enough time has passed, ω_B approaches some limit, the torque on drum B will approach zero, thus ω_B will approach some constant.]

But as long as you're satisfied that drum A continues at a constant angular velocity, you can use your previous work with conservation of momentum to determine the motion of drum B. Conservation of moment (as you've already done in your original post) makes this problem a lot easier.

This problem sure has a lot of physics in it.

I should say so.

 

[asteriatic]

Your approach is correct, but there is one error in your calculation. When you substitute λt=MB and b=2a, you should also take into account the fact that the sand is sticking to the outer drum. This means that the outer drum will experience an increase in mass as the sand accumulates on its surface. Therefore, the moment of inertia for the outer drum should be (MB+λt) instead of just MB. This will affect the final expression for ωB.

 

[asteriatic]

Your approach is correct, but there are a few mistakes in your equations.

Firstly, the moment of inertia for a solid cylinder is given by I = 1/2MR^2, not R^2Mω. So the correct equation for angular momentum would be L = 1/2(MA+MS)a^2ωA(0) and L = 1/2(MA+MS-λt)a^2ωA + 1/2(MB+λt)b^2ωB.

Secondly, in the conservation of angular momentum equation, you have used b = 2a, but this is only true if the two drums have the same mass. In this case, b > a, so you cannot make this substitution.

Lastly, when you substitute λt = MB, you are assuming that the sand has completely transferred from the inner drum to the outer drum. However, this is not the case as the sand is being transferred at a constant rate λ, so the sand in the inner drum is decreasing but not completely disappearing.

To solve this problem, you will need to set up and solve a differential equation for the angular velocities ωA and ωB. This will take into account the changing mass of the inner drum as the sand is transferred to the outer drum. You can then use the initial conditions (ωA(0) = ωA(0), ωB(0) = 0) to solve for the final angular velocities.