A block of mass m is kept on a disk rotating in the horizontal plane, 1 cm from the axis of rotation. When the disk rotates at a steady rate of 3.5 revolutions per seconds, the block slides on the disk, radially outwards, with an acceleration of 0.84 ` m/s^2` . Then find the coefficient of friction between the block and the disk. `(g = 10 m/s^2)`

Options

A. 0.2
B. 0.4
C. 0.6
D. 0.8

Answer

To find the coefficient of friction between the block and the disk, we can follow these steps:

### Step 1: Identify the given data
- Mass of the block: \( m \) (unknown)
- Distance from the axis of rotation: \( r = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Rate of rotation: \( f = 3.5 \, \text{revolutions/second} \)
- Acceleration of the block: \( a = 0.84 \, \text{m/s}^2 \)
- Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

### Step 2: Convert the rotational speed to linear velocity
The linear velocity \( v \) of the block can be calculated using the formula:
\[ v = 2 \pi r f \]
Substituting the values:
\[ v = 2 \pi (0.01) (3.5) = 0.07 \pi \, \text{m/s} \approx 0.22 \, \text{m/s} \]

### Step 3: Calculate the centrifugal force
The centrifugal force \( F \) acting on the block can be calculated using:
\[ F = \frac{mv^2}{r} \]
Substituting the values:
\[ F = \frac{m (0.22)^2}{0.01} = \frac{m \cdot 0.0484}{0.01} = 4.84m \, \text{N} \]

### Step 4: Set up the equation of motion
The net force acting on the block in the radial direction is given by:
\[ ma = F - f \]
Where \( f \) is the frictional force. Rearranging gives:
\[ ma + f = F \]
Substituting for \( F \):
\[ ma + f = 4.84m \]
Substituting \( a = 0.84 \, \text{m/s}^2 \):
\[ m(0.84) + f = 4.84m \]
This simplifies to:
\[ f = 4.84m - 0.84m = 4.00m \]

### Step 5: Calculate the coefficient of friction
The coefficient of friction \( \mu \) is given by:
\[ \mu = \frac{f}{N} \]
Where \( N \) is the normal force. The normal force \( N \) is equal to the weight of the block:
\[ N = mg \]
Substituting for \( f \):
\[ \mu = \frac{4.00m}{mg} = \frac{4.00}{10} = 0.4 \]

### Final Answer
The coefficient of friction between the block and the disk is \( \mu = 0.4 \).

---