1. Initial situation

A hoop of radius $r$ and mass $m$ is initially spinning with angular velocity $\omega_0$, but its center is at rest:

$v_{\text{CM}}(0)=0$

Since it is placed on a rough horizontal surface, friction acts at the point of contact.

• Initially, the bottom point of the hoop has speed $r\omega_0$ relative to the center.
• So the contact point slips on the ground.
• Friction opposes this slipping and therefore:
  • gives the center of mass a forward linear acceleration,
  • gives a retarding torque that reduces angular speed.

2. Friction force and linear acceleration

Let the friction force be $f$.

Then the translational equation is

$f = ma$

so

$a = \frac{f}{m}$

3. Torque and angular deceleration

For a hoop,

$I = mr^2$

Torque due to friction about the center:

$\tau = fr$

Thus angular deceleration is

$\alpha = \frac{\tau}{I} = \frac{fr}{mr^2} = \frac{f}{mr}$

This reduces the angular speed, so effectively

$\frac{d\omega}{dt} = -\frac{f}{mr}$

4. Relation between $v$ and $\omega$ during slipping

From translation:

$\frac{dv}{dt} = \frac{f}{m}$

From rotation:

$\frac{d\omega}{dt} = -\frac{f}{mr}$

Multiply the second equation by $r$:

$r\frac{d\omega}{dt} = -\frac{f}{m}$

Now add with the first equation:

$\frac{dv}{dt} + r\frac{d\omega}{dt} = 0$

So,

$\frac{d}{dt}(v + r\omega)=0$

Hence,

$v + r\omega = \text{constant}$

Initially,

$v=0, \quad \omega=\omega_0$

Therefore,

$v + r\omega = r\omega_0$

5. Condition when slipping ceases

The hoop stops slipping when pure rolling begins. Then

$v = r\omega$

Substitute into

$v + r\omega = r\omega_0$

Since $v=r\omega$, we get

$v + v = r\omega_0$

$2v = r\omega_0$

$v = \frac{r\omega_0}{2}$

6. Checking options

• A: $\dfrac{r\omega_0}{4}$
• B: $\dfrac{r\omega_0}{3}$
• C: $\dfrac{r\omega_0}{2}$
• D: $r\omega_0$

So the correct option is:

$\boxed{\text{C }\; \frac{r\omega_0}{2}}$