A coin placed on a rotating gramophone disc, remains at rest when it is at a distance of 9 cm from its centre ,the angular velocity of the disc is then tripled .At what distance from the centre ,the coin should be placed , so that it will remain at rest ?

Options

A. 2cm
B. 1 cm
C. 3 cm
D. 6 cm

Explanation

To solve the problem step-by-step, we need to analyze the forces acting on the coin placed on the rotating gramophone disc.

### Step 1: Understand the initial conditions
The coin is at rest at a distance of 9 cm from the center of the disc. This means that the centripetal force acting on the coin due to the rotation of the disc is balanced by the frictional force acting on the coin.

### Step 2: Write the equation for the initial condition
The centripetal force \( F_c \) required to keep the coin in circular motion is given by:
\[ F_c = m \omega^2 r \]
where:
- \( m \) is the mass of the coin,
- \( \omega \) is the angular velocity of the disc,
- \( r \) is the radius (distance from the center), which is 9 cm or 0.09 m.

Since the coin is at rest, the frictional force \( f \) must equal the centripetal force:
\[ f = m \omega^2 (0.09) \]

### Step 3: Analyze the new conditions after tripling the angular velocity
When the angular velocity is tripled, we have:
\[ \omega' = 3\omega \]
The new centripetal force required to keep the coin in circular motion at a new distance \( x \) from the center is:
\[ F_c' = m (3\omega)^2 x = 9m \omega^2 x \]

### Step 4: Set up the equation for the new condition
The frictional force remains the same since it depends on the mass of the coin and the coefficient of friction, which are unchanged. Therefore, we can equate the new centripetal force to the original frictional force:
\[ m \omega^2 (0.09) = 9m \omega^2 x \]

### Step 5: Simplify the equation
We can cancel \( m \) and \( \omega^2 \) from both sides (assuming they are non-zero):
\[ 0.09 = 9x \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{0.09}{9} = 0.01 \text{ m} = 1 \text{ cm} \]

### Conclusion
The coin should be placed at a distance of **1 cm** from the center of the disc to remain at rest after the angular velocity is tripled.

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