A uniform disc of radius `R` is spinned to the angular velocity `omega` and then carefully placed on a horizontal surface. How long will the disc be rotating on the surface if the friction coeffiecient is equal to `k`? The pressure exerted by the disc on the surface can be regarded as uniform.

Options

A. `(4omega R)/(3 mu g)`
B. `(3omega R)/(4 mug)`
C. `(omega R)/(3 mu g)`
D. `(4 omega R)/(mu g)`

Explanation

To solve the problem of how long a uniform disc of radius \( R \) spinning with an angular velocity \( \omega \) will continue to rotate on a horizontal surface with a friction coefficient \( k \), we can follow these steps:

### Step 1: Identify Forces Acting on the Disc
When the disc is placed on the surface, the forces acting on it include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.
- The frictional force \( F_f \) acting tangentially at the edge of the disc, opposing the motion.

Since there is no vertical acceleration, we can write:
\[ N = mg \]

### Step 2: Determine the Frictional Force
The frictional force can be expressed in terms of the coefficient of friction \( k \):
\[ F_f = kN = k(mg) \]

### Step 3: Calculate the Torque Due to Friction
The frictional force creates a torque \( \tau \) about the center of the disc. The torque due to the frictional force is given by:
\[ \tau = F_f \cdot R = k(mg) \cdot R \]

### Step 4: Relate Torque to Angular Acceleration
According to Newton's second law for rotation, the net torque is also related to angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the disc. For a uniform disc, the moment of inertia is:
\[ I = \frac{1}{2} m R^2 \]

Setting the two expressions for torque equal gives:
\[ k(mg)R = \left(\frac{1}{2} m R^2\right) \alpha \]

### Step 5: Solve for Angular Acceleration
We can simplify this equation to find \( \alpha \):
\[ \alpha = \frac{2k g}{R} \]

### Step 6: Use Angular Motion Equation
Since the disc starts with an initial angular velocity \( \omega \) and comes to rest, we can use the angular motion equation:
\[ \omega_f = \omega_i + \alpha t \]
where \( \omega_f = 0 \) (final angular velocity), \( \omega_i = \omega \) (initial angular velocity), and \( \alpha \) is negative (deceleration):
\[ 0 = \omega - \left(\frac{2k g}{R}\right) t \]

### Step 7: Solve for Time \( t \)
Rearranging the equation gives:
\[ t = \frac{\omega R}{2k g} \]

### Final Answer
Thus, the time for which the disc will continue to rotate on the surface is:
\[ t = \frac{\omega R}{2k g} \]

---