A block of mass m = 20 kg is kept is a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, `alpha = 3 rad//sec^(2)`. The friction coefficient between block and table is `mu = 0.5`. At time `t = (x)/(30)` from starting of motion (i.e. t =0) the block is just about to slip. Find the value of x `(g = 10 m//s^(2))`

Explanation

To solve the problem, we need to analyze the forces acting on the block on the rotating turntable. The block is subjected to centripetal and tangential forces due to the rotation of the table, and we need to find the time when the block is just about to slip.

### Step 1: Identify the given values
- Mass of the block, \( m = 20 \, \text{kg} \)
- Distance from the central axis, \( R = 1 \, \text{m} \)
- Angular acceleration, \( \alpha = 3 \, \text{rad/s}^2 \)
- Coefficient of friction, \( \mu = 0.5 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the normal force
The normal force \( N \) acting on the block is equal to its weight:
\[ N = mg = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Calculate the maximum frictional force
The maximum static frictional force \( F_f \) can be calculated using the coefficient of friction:
\[ F_f = \mu N = 0.5 \times 200 \, \text{N} = 100 \, \text{N} \]

### Step 4: Determine the angular velocity at time \( t \)
The angular velocity \( \omega \) at time \( t \) can be calculated using the formula:
\[ \omega = \alpha t = 3t \, \text{rad/s} \]

### Step 5: Calculate the centripetal force
The centripetal force \( F_c \) acting on the block is given by:
\[ F_c = m \frac{\omega^2 R}{R} = m \omega^2 = 20 \, \text{kg} \times (3t)^2 = 20 \times 9t^2 = 180t^2 \, \text{N} \]

### Step 6: Calculate the tangential force
The tangential force \( F_t \) is given by:
\[ F_t = m \alpha R = 20 \, \text{kg} \times 3 \, \text{rad/s}^2 \times 1 \, \text{m} = 60 \, \text{N} \]

### Step 7: Set up the equation for slipping condition
The block will just start to slip when the maximum frictional force is equal to the resultant of the centripetal and tangential forces:
\[ F_f = \sqrt{F_c^2 + F_t^2} \]
Substituting the values:
\[ 100 = \sqrt{(180t^2)^2 + (60)^2} \]

### Step 8: Square both sides to eliminate the square root
\[ 10000 = (180t^2)^2 + 3600 \]
\[ 10000 - 3600 = (180t^2)^2 \]
\[ 6400 = (180t^2)^2 \]

### Step 9: Solve for \( t^2 \)
Taking the square root:
\[ 80 = 180t^2 \]
\[ t^2 = \frac{80}{180} = \frac{8}{18} = \frac{4}{9} \]

### Step 10: Solve for \( t \)
\[ t = \sqrt{\frac{4}{9}} = \frac{2}{3} \, \text{s} \]

### Step 11: Relate \( t \) to \( x \)
Given that \( t = \frac{x}{30} \):
\[ \frac{2}{3} = \frac{x}{30} \]
Cross-multiplying gives:
\[ 2 \times 30 = 3x \]
\[ 60 = 3x \]
\[ x = 20 \]

### Final Answer
The value of \( x \) is \( 20 \).

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