A disc of radius 5 m is rotatating with angular frequency 10 rad/sec. A block of mass 2 kg is to be put on the disc frication coefficient between disc and block is`mu_(K)= 0.4`, then find the maximum distance from axis where the block can be placed without slidding:

Options

A. `2 cm`
B. `3 cm`
C. 4 cm
D. 6cm

Explanation

To solve the problem step by step, we need to determine the maximum distance from the axis of the rotating disc where the block can be placed without sliding off due to the centrifugal force acting on it.

### Step 1: Understanding the Forces
When the block is placed on the rotating disc, it experiences a centrifugal force due to the rotation. This force tries to push the block outward. The frictional force between the block and the disc must be sufficient to keep the block from sliding off.

### Step 2: Setting Up the Equation
The maximum frictional force \( F_f \) that can act on the block is given by:
\[ F_f = \mu_k \cdot m \cdot g \]
where:
- \( \mu_k = 0.4 \) (coefficient of kinetic friction),
- \( m = 2 \, \text{kg} \) (mass of the block),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

The centrifugal force \( F_c \) acting on the block when it is at a distance \( x \) from the axis is given by:
\[ F_c = m \cdot \omega^2 \cdot x \]
where \( \omega = 10 \, \text{rad/s} \) (angular frequency).

### Step 3: Equating Forces
For the block to not slide off, the frictional force must be equal to or greater than the centrifugal force:
\[ \mu_k \cdot m \cdot g \geq m \cdot \omega^2 \cdot x \]
We can simplify this by canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu_k \cdot g \geq \omega^2 \cdot x \]

### Step 4: Solving for \( x \)
Rearranging the equation to solve for \( x \):
\[ x \leq \frac{\mu_k \cdot g}{\omega^2} \]

### Step 5: Substituting Values
Now, substituting the known values into the equation:
\[ x \leq \frac{0.4 \cdot 9.8}{10^2} \]
Calculating the right-hand side:
\[ x \leq \frac{3.92}{100} = 0.0392 \, \text{m} \]
Converting to centimeters:
\[ x \leq 3.92 \, \text{cm} \]

### Step 6: Conclusion
The maximum distance from the axis where the block can be placed without sliding is approximately \( 3.92 \, \text{cm} \).