Unwrapping a rope out of a disk

Question

This exercise is inspired by exercises 83 and 100 of Chapter 10 in Giancoli's book

A uniform disk ($R = 0.85 m$; $M =21.0 kg$) has a rope wrapped around it. You apply a constant force $F = 35 N$ to unwrap it (at the point of contact ground-disk) while walking 5.5 m. Ignore friction.
a) How much has the center of mass of the disk moved? Explain.
Now derive a formula that relates the distance you have walked and how much rope has been unwrapped when:
b) You don't assume rolling without slipping.
c) You assume rolling without slipping.

a) I have two different answers here, which I guess one is wrong:
a.1) Here there's only one force to consider in the direction of motion: $\vec F$. Thus the center of mass should also move forward.
a.2) You are unwinding the rope out of the spool and thus exerting a torque $FR$ (I am taking counterclockwise as positive); the net force exerted on the CM is zero and thus the wheel only spins and the center of mass doesn't move.

The issue here is that my intuition tells me that there should only be spinning. I've been testing the idea with a paper roll and its CM does move forward, but I think this is due to the roll not being perfectly cylindrical; if the unwrapping paper were to be touching only at a point with an icy ground the CM's roll shouldn't move.

'What's your reasoning to assert that?'

Tangential velocity points forwards at distance $R$ below the disk's CM but this same tangential velocity points backwards at distance $R$ above the disk's CM and thus translational motion is cancelled out. Actually, we note that opposite points on the rim have opposite tangential velocities (assuming there's no friction so that the tangential velocity is constant).

My book assumes a.1) is OK. I say a.2) is OK. Who's right then?

b) We can calculate the unwrapped distance noting that the arc length is related to the radius by the angle (radian) enclosed:

$$\Delta s = R \Delta \theta$$

Assuming constant acceleration and zero initial angular velocity:

$$\Delta \theta = 1/2 \alpha t^2 = 1/2 \frac{\omega}{t} t^2 = 1/2 \omega t$$

By Newton's second Law (rotation) we can solve for $\omega$ and then plug it into the above equation:

$$\tau = FR = I \alpha = I \frac{\omega}{t} = 1/2 M R^2 \frac{\omega}{t}$$

$$\omega = \frac{2F}{M R}t$$

Let's plugg it into the other EQ.

$$\Delta \theta = \frac{F}{M R}t^2$$

Mmm we still have to eliminate $t$.

Assuming constant acceleration we get by the kinematic equation (note I am using the time $t$ you take to walk 5.5 m so that we know how much rope has been unwrapped in that time):

$$t^2 = \frac{2M\Delta x}{F}$$

Plugging it into $\Delta \theta$ equation:

$$\Delta \theta = \frac{2\Delta x}{R}$$

Plugging it into $\Delta s$ equation we get the equation we wanted:

$$\Delta s = 2 \Delta x$$

If we calculate both $v$ and $\omega$ we see that $v=R\omega$ is not true so the disk doesn't roll without slipping.

c) Here $v=R\omega$ must be true. We know that if that's the case the tangential velocity must be related to the center of mass' velocity as follows:

$$2v_{cm} = v$$

Assuming that the person holding the rope goes at speed $2v_{cm}$ we get:

$$\Delta x= 2 \Delta s$$

I get reversed equations at b) and c). How can we explain that difference in both equations beyond the fact of rolling without slipping?

Answer 1

Part a

Answer a.1 is correct.

Answer a.2 is incorrect because force and torque are not the same. There is only one force $F$ on the disk. This causes linear acceleration $a=F/M$ to the right say. Force $F$ does not pass through the CM of the disk so it creates a torque $\tau=FR$ about the CM, causing rotational acceleration $\alpha=\tau/J$ anti-clockwise, where $J$ is the moment of inertia about the CM.

Anti-clockwise torque $\tau$ does not cancel out the force $F$ to the right. These are different quantities so they cannot be added, just as mass and length cannot be added . The resultant force on the disk is not zero. Neither is the resultant torque.

Your intuition is incorrect. Your experiment is correct. But as you realise, it is not always like this. Sometimes the experiment is flawed. However, in this case it is your reasoning (intuition), to explain why the experiment failed, which is flawed.

You mention 3 possible explanations for the experiment not giving the result which you expect : (i) the disk is not perfectly cylindrical, (ii) there is friction between the disk and the ground (which could almost be eliminated if the experiment were performed on ice), and (iii) the velocities at opposite points on the rim of the disk are equal and opposite, so the midpoint between them (the CM) must be stationary.

I don't understand how you think explanation (i) might work. Perhaps it is related to (iii). Explanation (ii) works as follows :

The applied force $F$ acts to the right and a friction force $f$ acts to the left. When the experiment is performed on the ground then $f_g<F$. There is a resultant force $F-f_g$ on the disk to the right, so the CM of the disk moves to the right.

Next you argue that if we could make $f=0$ then the CM of the disk should stay where it is. However when we perform the experiment on ice, which causes less friction $f_i<f_g$, then we have $f_i<f_g<F$. The resultant force on the disk $F-f_i$ is still to the right but it is now greater than before, not less, because $f_i<f_g$. So reducing or removing friction (eg by performing the experiment on ice) makes the disk accelerate faster to the right. It does not make the CM accelerate slower or remain at rest.

Your argument in (iii) about the velocities on the rim of the disk cancelling out is also flawed. These two velocities are the same relative to the CM of the disk. They must be if the disk remains circular and does not break apart or change its shape. They could be different if the disk is not perfectly circular, as you suggest in explanation (i). Then they could be different distances from the CM. Nevertheless both points would have the same angular velocity about the CM.

However, the two points on the rim do not necessarily have the same velocities relative to the ground. So these two velocities do not have to be equal and opposite, and their vector sum (which gives the velocity of the CM) does not have to be zero. The disk does not have to remain stationary.

Parts b and c

I shall deal with these later.