A coin placed on a rotating turntable just slips if it is placed at a distance of 16 cm from the centre. If the angular velocity of the turnable is doubled, it will just slip at a distance ( in cm) of

Options

A. 2 cm
B. 3 cm
C. 4 cm
D. 5 cm

Explanation

To solve the problem, we need to analyze the forces acting on the coin placed on the rotating turntable and how they change when the angular velocity is doubled.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- When the coin is on the turntable, it experiences a centripetal force that keeps it moving in a circular path. This force is provided by friction between the coin and the turntable.
- The centripetal force required for circular motion is given by \( F_c = m \omega^2 r \), where \( m \) is the mass of the coin, \( \omega \) is the angular velocity, and \( r \) is the distance from the center of the turntable.

2. **Frictional Force**:
- The maximum static frictional force that can act on the coin before it slips is given by \( F_f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, \( N = mg \), where \( g \) is the acceleration due to gravity.
- Therefore, the frictional force can be expressed as \( F_f = \mu mg \).

3. **Condition for Slipping**:
- For the coin to just slip, the centripetal force must equal the maximum frictional force:
\[ \mu mg = m \omega^2 r \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \omega^2 r \]

4. **Initial Condition**:
- Initially, the coin slips at a distance \( r = 16 \) cm. Thus, we can write:
\[ \mu g = \omega^2 \times 16 \]

5. **Doubling the Angular Velocity**:
- Now, if the angular velocity is doubled, we have \( \omega' = 2\omega \). We need to find the new distance \( r' \) at which the coin will slip:
\[ \mu g = \omega'^2 r' \]
- Substituting \( \omega' = 2\omega \):
\[ \mu g = (2\omega)^2 r' = 4\omega^2 r' \]

6. **Setting Equations Equal**:
- We have two expressions for \( \mu g \):
\[ \omega^2 \times 16 = 4\omega^2 r' \]
- Dividing both sides by \( \omega^2 \) (assuming \( \omega \neq 0 \)):
\[ 16 = 4r' \]

7. **Solving for \( r' \)**:
- Rearranging the equation gives:
\[ r' = \frac{16}{4} = 4 \text{ cm} \]

### Final Answer:
The distance at which the coin will just slip when the angular velocity is doubled is **4 cm**.