MIT 8.01: Unwinding Drum Problem

[zenterix]

I'm self-studying MIT OCW's 8.01, Introduction to Classical Mechanics Course. I am on the final week, where the topic is translational and rotational motion. I was following along the course notes and reached an example which I'd like to dive a bit deeper into, but I am not sure how.
The problem is Example 20.4 in the notes if you want to see it directly. I will reproduce it here:

The example itself is only asking for the relationship between the angular accelerations and the linear acceleration of drum A. To find the answer it is straightforward:

The metal tape connecting the drums has a fixed total length. Whatever increase in the length between the drums is due to unrolling of A and unrolling of B. The increase in length between the drums is also the displacement of the center of mass of A. Therefore

$$\Delta y = R \Delta \theta_B + R \Delta \theta_A$$

Dividing by $\Delta t$ and taking the limit as $\Delta t \to 0$ we get

$$\frac{dy}{dt}=R(\alpha_A +\alpha_B)$$

That is intuitively clear, and that is the answer to this particular question. But what about finding the actual tension force or the angular accelerations of each drum?

In particular, the system of drums starts with zero momentum and angular momentum. There are no external torques, only the internal torque applied by the tape on each of the drums. The first question is: we need to calculate torques and angular momentums always relative to the same origin, correct?

If we use the center of mass of B as the origin, and call $l$ the length of the tape between the centers of the drums, we have:

$$\vec{\tau}_{cm_B,B} = RT \hat{k}$$
$$\vec{\tau}_{cm_B,A}=(R\hat{i} +l \hat{j} \times (-T \hat{j})=-RT \hat{k}$$

Where the notation $\vec{\tau}_{cm_B,B}$ means the torque about the $cm_B$ of drum B.

We can see the torques cancel out, ie net torque on the system is zero. This means angular momentum will be constant for the system of two drums.

The question is: to use conservation of momentum, is it really as hard as it seems; namely, I need to calculate angular momentum for drum A relative to the chosen origin, the center of mass of B. This seems like a messy calculation. Is the problem tractable with just Newtonian Mechanics methods?
Δy=RΔθA

 

[TSny]

$$\Delta y = R \Delta \theta_B + R \Delta \theta_A$$
Dividing by $\Delta t$ and taking the limit as $\Delta t \to 0$ we get $$\frac{dy}{dt}=R(\alpha_A +\alpha_B)$$

Shouldn't the $\alpha$'s be $\omega$'s? You'll need to take another time derivative to get an equation involving $\ddot y$, $\alpha_A$ and $\alpha_B$.

In particular, the system of drums starts with zero momentum and angular momentum. There are no external torques, only the internal torque applied by the tape on each of the drums. The first question is: we need to calculate torques and angular momentums always relative to the same origin, correct?

If you want to use the general relation $\large \vec \tau^{\rm ext} = \frac{d \vec L}{dt}$ for some chosen system, then the torques and angular momentum should be calculated relative to the same origin.

For the system consisting of both drums and the metal tape, and choosing the origin at the center of $B$, the net external torque will not be zero. (Consider the force of gravity acting on drum $A$.) So, angular momentum about this origin will not be conserved. But, you could set up $\large \vec \tau^{\rm ext} = \frac{d \vec L}{dt}$if you want.

It is not necessary to use angular momentum concepts for this problem. You should be able to solve it using $F = ma$ and $\tau = I \alpha$ applied to each drum separately along with your kinematic equation that relates $\ddot y$, $\alpha_A$, and $\alpha_B$.

 

[wrobel]

are the Lagrange equations forbidden?

 

[zenterix]

Shouldn't the α's be ω's? You'll need to take another time derivative to get an equation involving y¨, αA and αB.

You are right, it was a typo on my part. The correct equation is

$$\dot{y}=R(\omega_A+\omega_B)$$
$$\ddot{y}=R(\alpha_A+\alpha_B)$$

You are also correct that the gravitational force on drum A applies a torque to the system about the center of mass of B.

$$\vec{\tau}_{cm_B,B,f}=R\hat{i} \times T \hat{j} = RT \hat{k}$$
$$\vec{\tau}_{cm_B,A,f}=(2R \hat{i}+l\hat{j}) \times mg \hat{j}=(2Rmg-RT)\hat{k}$$
$$\implies \vec{\tau}_{cm_B,sys,f}= 2Rmg \hat{k}$$

So indeed the torque of the system about $cm_B$ is not zero.

Now I will try the approach you suggested as I understood it:

Drum B
$2^{nd}$ Law: $$mg+T-N=0$$
Torque: $$\vec{\tau}_{cm_B,B,f}=R\hat{i} \times T \hat{j} = RT \hat{k}$$
$$\implies RT=I_{cm_B,B}\alpha_B$$

Drum A
$2^{nd}$ Law: $$mg-T=ma_A$$
Torque: $$\vec{\tau}_{cm_A,A,f}=-R\hat{i} \times (-T \hat{j})=RT\hat{k}$$
$$\implies RT=I_{cm_A,A}\alpha_A$$

So all the equations together looks like:
$$a_A=R(\alpha_A+\alpha_B)\tag{1}$$
$$mg+T-N=0\tag{2}$$
$$RT=I_{cm_B,B}\alpha_B\tag{3}$$
$$RT=I_{cm_A,A}\alpha_A\tag{4}$$
$$mg-T=ma_A\tag{5}$$

Here we have five equations and five unknowns: $a_A$, $\alpha_A$, $\alpha_B$, $T$, $N$.

If we assume that the moments of inertia are the same and equal to $I$, from $(3)$ and $(4)$ we get that $$\alpha_A=\alpha_B=\alpha$$

$$\alpha=\frac{RT}{I}$$

Substitute in the equation for $T$ results in

$$T=\frac{mgI}{I+2R^2m}$$

$$\alpha=\frac{mgR}{I+2R^2m}$$

$$a_A=\frac{2mgR^2}{I+2R^2m}$$

$$N=\frac{2mg(I+R^2m)}{I+2R^2m}$$

So now let me investigate this result a bit.

If $I=\frac{mR^2}{2}$ then

$$T=\frac{mg}{5}$$

$$\alpha=\frac{2g}{5R}$$

$$a_A=\frac{4g}{5}$$

$$N=\frac{6mg}{5}$$
I found a few things interesting in this result.

1) When mass increases, the z component of angular velocity of the drums doesn't change, and neither does the acceleration of drum A. In fact, the latter acceleration is constant, given acceleration of gravity, no matter the other parameters. The tension force increases, but so does the moment of inertia of each drum. The tension increases just enough to keep the angular acceleration the same for the new mass.

2) The only variable that depends on the radius of the drums is z component of angular acceleration, $\alpha$. Moment of inertia increases with the square of radius. Torque increases linearly with radius on each drum, therefore the angular acceleration decreases. As radius keeps increasing, torque becomes insufficient to rotate each drum much in terms of angular displacement. The arc length rotated however seems to stay the same.

$$\alpha=\frac{2g}{5R} \implies \omega=\frac{2gt}{5R}\implies \theta=\frac{gt^2}{5R}$$

$$s(t)=R\theta(t) = \frac{gt^2}{5}$$

Indeed arc length as a function of time is the same for any radius.

 

[zenterix]

are the Lagrange equations forbidden?

Yes I don't know about that yet.

 

[TSny]

Very nice solution and analysis. If you like, you can check that $\large \vec \tau^{ext} = \frac{d \vec L}{dt}$ for the system as a whole with the origin chosen at the center of B. Here $\vec L$ is the total angular momentum of the system about the origin.

 

[zenterix]

1) When mass increases, the z component of angular velocity of the drums doesn't change, and neither does the acceleration of drum A. In fact, the latter acceleration is constant, given acceleration of gravity, no matter the other parameters. The tension force increases, but so does the moment of inertia of each drum. The tension increases just enough to keep the angular acceleration the same for the new mass.

For the record, I wrote "angular velocity" but I meant "angular acceleration" of the drums doesn't change.

 

[zenterix]

We can now check the angular momentum of the system

$$\vec{L}_{cm_B,B}=\frac{mR^2}{2}\frac{2gt}{5R}\hat{k}=\frac{Rgmt}{5}\hat{k}\tag{1}$$

$$\vec{L}_{cm_B,A}=\vec{L}_{cm_B, cm_A} +\vec{L}_{cm_A,A}\tag{2}$$

In $(2)$ I used the relationship that says that the angular momentum of a system of particles about $cm_B$ is the angular moment of the center of mass of the system about $cm_B$ plus the angular momentum of the particles about the center of mass of the system of particles. In this case the system of particles is the rigid body drum A.

$$\implies \vec{L}_{cm_B,A}=(2R\hat{i}+l \hat{j}) \times m \frac{4gt}{5} \hat{j} + \frac{mR^2}{2} \frac{2gt}{5R} \hat{k}=\frac{Rgmt}{5} \hat{k}$$

$$\implies \vec{L}_{cm_B,sys}=2Rgmt \hat{k}$$

$$\frac{d\vec{L}_{cm_B}}{dt}=2Rgmt \hat{k} = \vec{\tau}_{cm_B,sys}$$

This value of $\vec{\tau}_{cm_B,sys}$ agrees with the value calculated initially by calculating the torque on each individual drum.

 

[TSny]

That looks very good. I noticed a couple of typographical errors:

$$\implies \vec{L}_{cm_B,A}=(2R\hat{i}+l \hat{j}) \times m \frac{4gt}{5} \hat{j} + \frac{mR^2}{2} \frac{2gt}{5R} \hat{k}=\frac{Rgmt}{5} \hat{k}$$

$$\implies \vec{L}_{cm_B,sys}=2Rgmt \hat{k}$$

In the first equation above, the far right-hand side is missing a factor of 9. But the second equation is correct.

$$\frac{d\vec{L}_{cm_B}}{dt}=2Rgmt \hat{k} = \vec{\tau}_{cm_B,sys}$$

The $t$ in the expression after the first equal sign should not be there.

This value of $\vec{\tau}_{cm_B,sys}$ agrees with the value calculated initially by calculating the torque on each individual drum.

Good. You only need to consider external torques acting on the system. The only external force that contributes a torque about the cm of B is the force of gravity on A. This force has a lever arm of 2R. So, $\vec{\tau}_{cm_B,sys} = 2Rmg \hat{k}$

 

[wrobel]

f you like, you can check that for the system as a whole with the origin chosen at the center of B.

and to obtain a second dynamical equation one can differentiate the energy conservation law in time
the both equations are free from the tension forces