A coin kept on a rotating gramophone disc just beging to slip if its centre is at a distance of 8 cm from the centre of the disc .The angular velocity of the gramophone disc is then doubled , Through what distance , the coin should be shifted towards the centre , so that the coin will just slip ?

Options

A. 2 cm
B. 4 cm
C. 6 cm
D. 16 cm

Explanation

To solve the problem step by step, we will analyze the forces acting on the coin when it is on the rotating gramophone disc.

### Step 1: Understand the Initial Condition
The coin is initially at a distance \( r = 8 \, \text{cm} \) from the center of the disc. It is just about to slip, meaning the frictional force equals the required centripetal force for circular motion.

### Step 2: Write the Equation for the Initial Condition
The centripetal force required to keep the coin moving in a circle 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 radius (8 cm in this case).

The frictional force \( f \) that prevents the coin from slipping is given by:
\[ f = \mu m g \]
where \( \mu \) is the coefficient of static friction and \( g \) is the acceleration due to gravity.

Since the coin is just about to slip, we have:
\[ \mu m g = m \omega^2 r \]
Cancelling \( m \) from both sides, we get:
\[ \mu g = \omega^2 r \]

### Step 3: Analyze the New Condition
When the angular velocity is doubled, the new angular velocity \( \omega' = 2\omega \). The coin will now slip at a new radius \( r' \).

### Step 4: Write the Equation for the New Condition
The centripetal force in the new condition can be expressed as:
\[ F_c' = m \omega'^2 r' = m (2\omega)^2 r' = 4m \omega^2 r' \]
Setting this equal to the frictional force, we have:
\[ \mu m g = 4m \omega^2 r' \]
Again cancelling \( m \), we get:
\[ \mu g = 4 \omega^2 r' \]

### Step 5: Relate the Two Conditions
From the first condition, we have:
\[ \mu g = \omega^2 r \]
Substituting \( r = 8 \, \text{cm} \):
\[ \mu g = \omega^2 \cdot 8 \]

Now substituting this into the second condition:
\[ 8 = 4 r' \]
This implies:
\[ r' = \frac{8}{4} = 2 \, \text{cm} \]

### Step 6: Calculate the Shift Towards the Center
The coin needs to be shifted from \( r = 8 \, \text{cm} \) to \( r' = 2 \, \text{cm} \). The distance \( x \) that the coin should be shifted is:
\[ x = r - r' = 8 \, \text{cm} - 2 \, \text{cm} = 6 \, \text{cm} \]

### Final Answer
The coin should be shifted **6 cm** towards the center of the disc to just begin to slip again.

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