This problem involves a hoop placed on a rough horizontal surface. Initially, it has an angular velocity ${\omega }_0$ but its center of mass has zero linear velocity. Friction acts to oppose slipping, eventually causing the hoop to roll without slipping. We need to find the center of mass velocity at that point.

Key Concepts:

1. Condition for Pure Rolling: When an object rolls without slipping, the linear velocity of its center of mass (v) is related to its angular velocity (ω) by the equation: $v=r\omega$.

2. Role of Friction: The frictional force acts at the point of contact. It opposes the relative motion (slipping) and does two things:

• It provides a linear acceleration (a) to the center of mass.
• It provides a torque (τ) that causes an angular deceleration (α).

3. Equations of Motion:
Let the frictional force be $f$.
Linear Motion: $f=ma$ (Newton's Second Law)
Rotational Motion: The torque due to friction is $\tau =fr$. This torque causes an angular acceleration. For a hoop, the moment of inertia about its center is $I=m{r}^2$.
Using $\tau =I\alpha$, we get: $fr=m{r}^2\alpha$, which simplifies to $f=mr\alpha$.

4. Kinematics: Let's analyze how velocity and angular velocity change with time until slipping stops.
The linear velocity of the center of mass increases from 0: $v=0+at=at$.
The angular velocity decreases from ${\omega }_0$: $\omega ={\omega }_0-\alpha t$ (friction acts to slow down the rotation).

5. Solving for the Final Condition: Slipping ceases when $v=r\omega$.
Substitute the kinematic equations into the pure rolling condition: $at=r({\omega }_0-\alpha t)$
From our force equations, we found $f=ma$ and $f=mr\alpha$. This means $ma=mr\alpha$, which simplifies to a very important relation: $a=r\alpha$.
Substitute $\alpha =\frac{a}{r}$ into the pure rolling equation: $at=r({\omega }_0-\frac{a}{r}t)$
Simplify: $at=r{\omega }_0-at$
Bring like terms together: $at+at=r{\omega }_0$
This gives: $2at=r{\omega }_0$
Recall that the final velocity $v=at$. Therefore: $2v=r{\omega }_0$
Solving for $v$: $v=\frac{r{\omega }_0}{2}$

Final Answer: The velocity of the centre of the hoop when it ceases to slip is $\frac{r{\omega }_0}{2}$.

Related Topics and Formulae

Rotational Dynamics: This is the study of motion involving rotation, governed by Newton's laws for rotation. The key equation is $\tau =I\alpha$, which is the rotational analog of $F=ma$.

Pure Rolling (Rolling without Slipping): This is a combined motion of translation and rotation where there is no relative motion between the object and the surface at the point of contact. The fundamental constraint is $v=r\omega$.

Moment of Inertia (I): This is a measure of an object's resistance to changes in its rotation. It depends on the mass distribution relative to the axis of rotation. For common shapes:

• Hoop about central axis: $I=m{r}^2$
• Solid disk/cylinder about central axis: $I=\frac{1}{2}m{r}^2$
• Solid sphere about diameter: $I=\frac{2}{5}m{r}^2$

Kinetic Energy in Rolling: The total kinetic energy (K) of a rolling object is the sum of its translational and rotational kinetic energy: $K=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$.