A hoop of radius `r` 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 cases to slip ?

Options

A. `(romega_(0))/(3)`
B. `(romega_(0))/(2)`
C. `romega_(0)`
D. `(romega_(0))/(4)`

Answer

To solve the problem, we need to analyze the motion of the hoop as it transitions from slipping to rolling without slipping. Here’s the step-by-step solution:

### Step 1: Understand the Initial Conditions
The hoop has a radius \( r \) and mass \( m \). It is initially rotating with an angular velocity \( \omega_0 \) and has a center of mass velocity of \( 0 \).

### Step 2: Determine the Velocity of the Contact Point
The point of contact with the ground (let's call it point B) has a velocity due to the rotation of the hoop. The velocity of point B is given by:
\[ v_B = \omega_0 \cdot r \]
Since the hoop is initially at rest in terms of translational motion, this point is moving backward relative to the ground.

### Step 3: Analyze the Effect of Friction
As the hoop rotates, the frictional force acts at point B in the forward direction (to the right). This friction will reduce the angular velocity \( \omega_0 \) and create a linear velocity \( v \) in the forward direction.

### Step 4: Set Up the Condition for Rolling Without Slipping
For the hoop to cease slipping, the velocity of the center of mass \( v \) must equal the velocity of the point of contact B when it is at rest (which is zero). The net velocity of point A (the center of the hoop) can be expressed as:
\[ v_A = v - \omega r \]
Setting this equal to zero for rolling without slipping gives:
\[ v - \omega r = 0 \quad \Rightarrow \quad v = \omega r \]

### Step 5: Apply Conservation of Angular Momentum
The torque about point B is zero (since it is the point of contact), which means angular momentum is conserved about this point. The initial angular momentum \( L_i \) is:
\[ L_i = I \cdot \omega_0 = m r^2 \cdot \omega_0 \]
where \( I \) is the moment of inertia of the hoop about its center.

The final angular momentum \( L_f \) when the hoop is rolling without slipping is:
\[ L_f = I \cdot \omega + m v r = m r^2 \cdot \omega + m v r \]

### Step 6: Set Initial and Final Angular Momentum Equal
Setting \( L_i = L_f \):
\[ m r^2 \cdot \omega_0 = m r^2 \cdot \omega + m v r \]
Canceling \( m \) from both sides:
\[ r^2 \cdot \omega_0 = r^2 \cdot \omega + v r \]

### Step 7: Substitute \( v = \omega r \)
Substituting \( v = \omega r \) into the equation gives:
\[ r^2 \cdot \omega_0 = r^2 \cdot \omega + \omega r^2 \]
This simplifies to:
\[ r^2 \cdot \omega_0 = r^2 \cdot \omega + r^2 \cdot \omega \]
\[ r^2 \cdot \omega_0 = 2 r^2 \cdot \omega \]

### Step 8: Solve for \( v \)
Dividing both sides by \( 2 r^2 \) gives:
\[ \omega = \frac{\omega_0}{2} \]
Substituting back to find \( v \):
\[ v = \omega r = \frac{\omega_0}{2} \cdot r \]

### Final Result
Thus, the velocity of the center of the hoop when it ceases to slip is:
\[ v = \frac{\omega_0 r}{2} \]