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?**