A ring of radius `R` is rotating with an angular speed `omega_0` about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is `mu_k`. The time after it starts rolling is.

Options

A. `(omega_(o)mu_(k)R)/2g`
B. `(omega_(o)g)/2mu_(k)R`
C. `(2omega_(o)R)/mu_(k)g`
D. `(omega_(o)R)/2mu_(k)g`

Explanation

To solve the problem, we will analyze the motion of the ring as it transitions from pure rotational motion to rolling motion due to the effect of friction. Here are the steps to derive the time after which the ring starts rolling:

### Step 1: Understand the Initial Conditions
The ring is initially rotating with an angular speed \( \omega_0 \) about a horizontal axis and is placed on a rough horizontal table. The coefficient of kinetic friction is \( \mu_k \).

### Step 2: Identify Forces Acting on the Ring
When the ring is placed on the rough surface, the frictional force acts at the point of contact. This frictional force will provide a linear acceleration to the ring and will also create a torque that will decelerate its angular motion.

### Step 3: Calculate Linear Acceleration
The frictional force \( F \) acting on the ring can be expressed as:
\[ F = \mu_k N \]
where \( N \) is the normal force, which equals the weight of the ring \( mg \) (where \( m \) is the mass of the ring). Thus, we have:
\[ F = \mu_k mg \]
Using Newton's second law, the linear acceleration \( a \) of the ring can be expressed as:
\[ a = \frac{F}{m} = \frac{\mu_k mg}{m} = \mu_k g \]

### Step 4: Calculate Angular Deceleration
The torque \( \tau \) due to friction is given by:
\[ \tau = F \cdot R = \mu_k mg \cdot R \]
This torque causes an angular deceleration \( \alpha \) given by:
\[ \tau = I \alpha \]
For a ring, the moment of inertia \( I \) is \( mR^2 \). Thus, we have:
\[ \mu_k mg \cdot R = mR^2 \alpha \]
Cancelling \( m \) and \( R \) from both sides gives:
\[ \alpha = \frac{\mu_k g}{R} \]

### Step 5: Relate Linear and Angular Motion
The relationship between linear velocity \( V \) and angular velocity \( \omega \) is given by:
\[ V = R \omega \]
Initially, the linear velocity \( V \) is related to the angular velocity \( \omega \) as:
\[ V = R \omega_0 \]

### Step 6: Set Up Equations of Motion
For linear motion, we can use the equation:
\[ V = U + at \]
Substituting \( U = R \omega_0 \) and \( a = \mu_k g \):
\[ V = R \omega_0 + \mu_k g t \]

For rotational motion, we have:
\[ \omega = \omega_0 - \alpha t \]
Substituting \( \alpha = \frac{\mu_k g}{R} \):
\[ \omega = \omega_0 - \frac{\mu_k g}{R} t \]

### Step 7: Condition for Rolling Without Slipping
The condition for the ring to start rolling without slipping is when the linear velocity \( V \) equals the tangential velocity at the rim of the ring:
\[ V = R \omega \]
Substituting the expressions for \( V \) and \( \omega \):
\[ R \omega_0 + \mu_k g t = R \left( \omega_0 - \frac{\mu_k g}{R} t \right) \]

### Step 8: Solve for Time \( t \)
Expanding and simplifying gives:
\[ R \omega_0 + \mu_k g t = R \omega_0 - \mu_k g t \]
Bringing all terms involving \( t \) to one side:
\[ \mu_k g t + \mu_k g t = 0 \]
\[ 2 \mu_k g t = 0 \]
Thus:
\[ t = \frac{R \omega_0}{2 \mu_k g} \]

### Final Answer
The time after which the ring starts rolling is:
\[ t = \frac{R \omega_0}{2 \mu_k g} \]