A hoop of radius r and mass m rotating with an angular velocity $$\omega_0$$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

We have a thin hoop of mass $$m$$ and radius $$r$$ whose centre is initially at rest, $$v_0=0$$. However, the hoop is already spinning about its central axis with angular velocity $$\omega_0$$. The hoop is now gently placed on a rough horizontal surface. Because the point of the hoop that touches the ground is sliding with respect to the ground, a kinetic friction force begins to act.

The direction of the initial slipping decides the direction of friction. If we imagine that the hoop is rotating clockwise, the lowest point of the rim is trying to move towards the left with speed $$r\omega_0$$. The rough ground therefore exerts a friction force $$f$$ towards the right on the hoop (opposite to the relative motion). This single horizontal force does two things simultaneously:

1. It gives the whole hoop a forward linear impulse, so the centre of mass acquires some velocity $$v$$.
2. It produces a torque about the centre that opposes the original spin and therefore reduces the angular velocity from $$\omega_0$$ to some smaller value $$\omega$$.

We now write the impulse-momentum relations algebraically and keep every step explicit.

Let the total time during which slipping takes place be $$t$$ and let the (constant) kinetic friction force be $$f$$. The linear impulse delivered to the hoop is $$J=\displaystyle\int_0^{t}f\,dt = f t$$. The linear momentum gained by the hoop is $$m v$$. Using the linear impulse-momentum theorem

$$\text{Linear impulse} = \text{Change in linear momentum},$$

we can write

$$f t = m v \;. \quad -(1)$$

Next we use the angular impulse-momentum theorem about the centre of the hoop. The torque produced by friction about the centre is

$$\tau = f r,$$

and it acts opposite to the initial spin. The angular impulse about the centre is therefore $$-\tau t = -f r t$$ (negative sign because it reduces $$\omega$$). The change in angular momentum about the centre is $$I(\omega-\omega_0)$$ where, for a thin hoop,

$$I = m r^2.$$

Writing the angular impulse-momentum theorem,

$$-f r t = m r^2 (\omega-\omega_0). \quad -(2)$$

We now substitute $$f t$$ from equation (1) into equation (2). From (1) we have $$f t = m v$$, so (2) becomes

$$-r(m v) = m r^2 (\omega-\omega_0).$$

Dividing both sides by $$m r$$ (since $$m\neq0, r\neq0$$), we obtain

$$-v = r(\omega-\omega_0),$$

or after rearranging,

$$\omega = \omega_0 - \frac{v}{r}. \quad -(3)$$

The hoop stops slipping at the instant pure rolling sets in. The pure rolling (no-slip) condition is

$$v = r\omega. \quad -(4)$$

We now substitute the expression for $$\omega$$ from equation (3) into the rolling condition (4).

Putting (3) into (4) gives

$$v = r\left(\omega_0 - \frac{v}{r}\right).$$

Simplify the right-hand side step by step:

$$v = r\omega_0 - v.$$

Now gather the like terms:

$$v + v = r\omega_0,$$

$$2v = r\omega_0.$$

Finally divide by 2 to isolate $$v$$:

$$v = \frac{r\omega_0}{2}.$$ \quad -(5)

Thus, when slipping just ceases and pure rolling commences, the centre of the hoop moves forward with speed $$\displaystyle \frac{r\omega_0}{2}$$.

Hence, the correct answer is Option A.