A block of mass m is kept on rough horizontal tum table at a distance r from centre of table. Coefficient of friction between turn table and block is `mu`. Now turn table starts rotating with uniform angular acceleration `alpha`.
Find angle made by friction force with velocity at the point of slipping,

Options

A. `theta = tan^(-1)(alphat)`
B. `theta = tan^(-1)(alphat^(2))`
C. `theta = tan^(-1)(alphat^(3))`
D. None of above

Answer

To solve the problem step by step, let's break down the concepts and calculations involved.

### Step 1: Understand the System
We have a block of mass \( m \) placed on a rough horizontal turntable at a distance \( r \) from the center. The turntable is rotating with a uniform angular acceleration \( \alpha \). The coefficient of friction between the block and the turntable is \( \mu \).

### Step 2: Identify the Forces Acting on the Block
When the turntable starts rotating, two types of accelerations act on the block:
1. **Tangential Acceleration (\( a_t \))**: Due to the angular acceleration of the turntable.
2. **Centripetal Acceleration (\( a_c \))**: Required to keep the block moving in a circular path.

### Step 3: Calculate Tangential Acceleration
The tangential acceleration \( a_t \) can be calculated using the formula:
\[ a_t = \alpha \cdot r \]

### Step 4: Calculate the Velocity of the Block
The velocity \( v \) of the block after time \( t \) can be derived from the kinematic equation:
\[ v = u + a_t \cdot t \]
Since the initial velocity \( u = 0 \):
\[ v = 0 + (\alpha \cdot r) \cdot t = \alpha \cdot r \cdot t \]

### Step 5: Calculate Centripetal Acceleration
The centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the expression for \( v \):
\[ a_c = \frac{(\alpha \cdot r \cdot t)^2}{r} = \alpha^2 \cdot r \cdot t^2 \]

### Step 6: Determine the Angle Between Friction and Velocity
At the point of slipping, the frictional force acts towards the center (radially inward), while the velocity of the block is tangential. The angle \( \theta \) between the frictional force and the velocity can be determined using the relationship between the centripetal and tangential accelerations.

Using the relationship:
\[ \tan(\theta) = \frac{a_c}{a_t} \]
Substituting the values we calculated:
\[ \tan(\theta) = \frac{\alpha^2 \cdot r \cdot t^2}{\alpha \cdot r} = \alpha \cdot t \]

### Step 7: Solve for the Angle
Thus, the angle \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}(\alpha \cdot t) \]

### Conclusion
The angle made by the friction force with the velocity at the point of slipping is:
\[ \theta = \tan^{-1}(\alpha \cdot t) \]