A disc of radius `R=2m` starts rotating wth constant angular acceleration `alpha=(2rad)//s^(2)`. A block of mass `m=2kg` is kept at a distance `(R)/(2)` from the centre of disc. The coeffiecient of friction between disc and block is `mu_(s)=0.4,mu_(k)=0.3`. Acceleration of block when it just slips w.r.t. ground and w.r.t. disc are respectively`-`(`g=10ms^(-2)`)

Options

A. `3ms^(-2),1ms^(-2)`
B. `1ms^(-2),3ms^(-2)`
C. `4ms^(-2),1ms^(-2)`
D. `4ms^(-2),3ms^(-2)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and apply the concepts of friction and circular motion.

### Step 1: Identify the given parameters
- Radius of the disc, \( R = 2 \, \text{m} \)
- Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
- Mass of the block, \( m = 2 \, \text{kg} \)
- Distance of the block from the center of the disc, \( r = \frac{R}{2} = 1 \, \text{m} \)
- Coefficient of static friction, \( \mu_s = 0.4 \)
- Coefficient of kinetic friction, \( \mu_k = 0.3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the maximum static friction force
The maximum static friction force \( F_s \) that can act on the block is given by:
\[ F_s = \mu_s \cdot N \]
where \( N \) is the normal force. The normal force on the block is equal to its weight since there are no vertical accelerations:
\[ N = m \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]
Thus, the maximum static friction force is:
\[ F_s = \mu_s \cdot N = 0.4 \cdot 20 \, \text{N} = 8 \, \text{N} \]

### Step 3: Calculate the required centripetal force
The block is at a distance \( r = 1 \, \text{m} \) from the center of the disc. The centripetal force \( F_c \) required to keep the block moving in a circle is given by:
\[ F_c = m \cdot a_c \]
where \( a_c \) is the centripetal acceleration, which can be expressed in terms of angular acceleration:
\[ a_c = r \cdot \alpha = 1 \, \text{m} \cdot 2 \, \text{rad/s}^2 = 2 \, \text{m/s}^2 \]
Thus, the required centripetal force is:
\[ F_c = m \cdot a_c = 2 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 4 \, \text{N} \]

### Step 4: Determine if the block slips
The block will slip when the required centripetal force exceeds the maximum static friction force:
\[ F_c > F_s \implies 4 \, \text{N} > 8 \, \text{N} \]
Since this is not true, the block will not slip due to static friction until the angular velocity increases.

### Step 5: Calculate the acceleration of the block when it just slips
When the block starts to slip, the kinetic friction force \( F_k \) will act on it:
\[ F_k = \mu_k \cdot N = 0.3 \cdot 20 \, \text{N} = 6 \, \text{N} \]
The net force acting on the block when it slips is:
\[ F_{\text{net}} = F_k - F_c = 6 \, \text{N} - 4 \, \text{N} = 2 \, \text{N} \]
Using Newton's second law, we can find the acceleration \( a \) of the block:
\[ F_{\text{net}} = m \cdot a \implies 2 \, \text{N} = 2 \, \text{kg} \cdot a \implies a = 1 \, \text{m/s}^2 \]

### Step 6: Calculate the acceleration of the block with respect to the ground
The acceleration of the block with respect to the ground when it just slips is the sum of the acceleration due to kinetic friction and the centripetal acceleration:
\[ a_{\text{ground}} = a + a_c = 1 \, \text{m/s}^2 + 2 \, \text{m/s}^2 = 3 \, \text{m/s}^2 \]

### Final Answers
- Acceleration of the block when it just slips with respect to the ground: \( 3 \, \text{m/s}^2 \)
- Acceleration of the block when it just slips with respect to the disc: \( 1 \, \text{m/s}^2 \)