A ring of radius `R` is first rotated with an angular velocity `omega` and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is `mu`. Time after which its angular speed is reduced to half is

Options

A. `(omega_(0)muR)/(2g)`
B. `(omega_(0)g)/(2muR)`
C. `(2omega_(0)R)/(mug)`
D. `(omega_(0)R)/(2mug)`

Answer

To solve the problem step by step, we will analyze the situation involving a ring rotating on a rough surface and determine the time it takes for its angular speed to reduce to half.

### Step 1: Understand the Forces Acting on the Ring
When the ring is placed on the rough surface, the frictional force acts to oppose the motion of the ring. The forces acting on the ring are:
- The gravitational force (weight) acting downward, \( mg \).
- The normal force \( N \) acting upward, which balances the weight, so \( N = mg \).
- The frictional force \( f \) acting opposite to the direction of rotation.

### Step 2: Calculate the Torque Due to Friction
The torque \( \tau \) acting on the ring due to the frictional force can be expressed as:
\[ \tau = f \cdot R \]
where \( f \) is the frictional force and \( R \) is the radius of the ring.

The maximum static frictional force is given by:
\[ f = \mu N = \mu mg \]
Thus, the torque becomes:
\[ \tau = \mu mg R \]

### Step 3: Relate Torque to Angular Acceleration
According to Newton's second law for rotation, the torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
For a ring, the moment of inertia \( I \) is given by:
\[ I = m R^2 \]
Substituting this into the torque equation gives:
\[ \mu mg R = m R^2 \alpha \]

### Step 4: Solve for Angular Acceleration
We can cancel \( m \) and \( R \) from both sides (assuming \( R \neq 0 \)):
\[ \mu g = R \alpha \quad \Rightarrow \quad \alpha = \frac{\mu g}{R} \]

### Step 5: Set Up the Equation for Angular Velocity
The angular velocity of the ring changes with time according to the equation:
\[ \omega = \omega_0 + \alpha t \]
where \( \omega_0 \) is the initial angular velocity and \( \alpha \) is the angular acceleration.

### Step 6: Find the Time When Angular Velocity is Halved
We need to find the time \( t \) when the angular velocity is reduced to half:
\[ \frac{\omega_0}{2} = \omega_0 - \alpha t \]
Rearranging gives:
\[ \frac{\omega_0}{2} = \omega_0 - \frac{\mu g}{R} t \]
\[ \frac{\omega_0}{2} - \omega_0 = -\frac{\mu g}{R} t \]
\[ -\frac{\omega_0}{2} = -\frac{\mu g}{R} t \]
\[ t = \frac{\omega_0 R}{2 \mu g} \]

### Final Answer
The time after which the angular speed of the ring is reduced to half is:
\[ t = \frac{\omega_0 R}{2 \mu g} \]

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