A small coin is placed at a distance r from the centre of a gramophone record.The rotational speed of the record is gradually increased. If the coefficient of friction between the coin and the record is `mu,` the minimum angular frequency of the record for which the coin will fly off is given by :

Options

A. `sqrt((2mug)/r)`
B. `sqrt((mug)/(2r))`
C. `sqrt((mug)/(r))`
D. `2sqrt((mug)/(r))`

Answer

To solve the problem of determining the minimum angular frequency of a gramophone record at which a coin will fly off, we can follow these steps:

### Step 1: Identify the Forces Acting on the Coin
When the gramophone record rotates, the coin experiences two main forces:
1. **Centrifugal Force**: This force acts outward and is given by the formula:
\[ F_{\text{centrifugal}} = m r \omega^2 \]
where \( m \) is the mass of the coin, \( r \) is the distance from the center of the record, and \( \omega \) is the angular frequency of the record.

2. **Frictional Force**: This force acts inward and is given by:
\[ F_{\text{friction}} = \mu N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force acting on the coin.

### Step 2: Establish the Relationship Between Forces
For the coin to remain on the record, the frictional force must be equal to or greater than the centrifugal force. At the point of flying off, these forces are equal:
\[ F_{\text{friction}} = F_{\text{centrifugal}} \]
Substituting the expressions for these forces, we get:
\[ \mu N = m r \omega^2 \]

### Step 3: Determine the Normal Force
The normal force \( N \) acting on the coin is equal to the weight of the coin, which can be expressed as:
\[ N = mg \]
where \( g \) is the acceleration due to gravity.

### Step 4: Substitute the Normal Force into the Equation
Substituting \( N = mg \) into the force balance equation gives:
\[ \mu (mg) = m r \omega^2 \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ \mu g = r \omega^2 \]

### Step 6: Solve for Angular Frequency \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega^2 = \frac{\mu g}{r} \]
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{\mu g}{r}} \]

### Final Answer
Thus, the minimum angular frequency of the record for which the coin will fly off is:
\[ \omega = \sqrt{\frac{\mu g}{r}} \]