A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of

Options

A. 1cm
B. 2cm
C. 4cm
D. 8cm

Answer

To solve the problem step by step, we will 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. **Understand the Forces Acting on the Coin:**
- When the coin is placed on the turntable, it experiences a centrifugal force due to the rotation, which is given by the formula:
\[ F_c = m r \omega^2 \]
- Here, \( m \) is the mass of the coin, \( r \) is the distance from the center of the turntable, and \( \omega \) is the angular velocity of the turntable.
- The centrifugal force is counteracted by the frictional force, which can be expressed as:
\[ F_f = \mu_s N \]
- Where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force (which equals \( mg \) for horizontal surfaces).

2. **Set Up the Equation for the Initial Condition:**
- At the initial distance \( r = 4 \) cm, the forces are balanced:
\[ m r \omega^2 = \mu_s mg \]
- Simplifying this by canceling \( m \) from both sides gives:
\[ r \omega^2 = \mu_s g \]
- This equation shows that the product \( r \omega^2 \) is a constant.

3. **Consider the New Condition When Angular Velocity is Doubled:**
- When the angular velocity is doubled, we have \( \omega' = 2\omega \).
- We need to find the new distance \( r' \) at which the coin will just slip:
\[ m r' (2\omega)^2 = \mu_s mg \]
- This simplifies to:
\[ r' \cdot 4\omega^2 = \mu_s g \]

4. **Equate the Two Conditions:**
- Since \( r \omega^2 = \mu_s g \) is constant, we can equate the two conditions:
\[ r \omega^2 = r' \cdot 4\omega^2 \]
- Canceling \( \omega^2 \) from both sides gives:
\[ r = 4 r' \]

5. **Solve for the New Distance \( r' \):**
- We know that \( r = 4 \) cm, so substituting this into the equation gives:
\[ 4 = 4 r' \]
- Dividing both sides by 4 results in:
\[ r' = 1 \text{ cm} \]

### Final Answer:
The distance at which the coin will just slip when the angular velocity is doubled is **1 cm**.