A coin is placed on the horizontal surface of a rotation disc. The distance of the coin from the axis is 1 m and coefficient of friction is 0.5. If the disc starts from rest and is given an angular acceleration `(1)/(sqrt(2))` rad /`sec^(2)`, the number of revolutions through which the disc turns before the coin slips is

Options

A. `(4)/(pi)`
B. `(7)/(4pi)`
C. `(5sqrt(2))/(4pi)`
D. `(3)/(4pi)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript and derive the necessary equations.

### Step 1: Identify the forces acting on the coin
The coin experiences two main forces:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The normal force \( N \) acting upwards.

Since the coin is on a horizontal surface, these forces balance each other:
\[ N = mg \]

### Step 2: Determine the centripetal force required to keep the coin moving in a circle
The required centripetal force \( F_c \) for the coin to stay on the disc is provided by the frictional force. The centripetal force can be expressed as:
\[ F_c = m \omega^2 R \]
where \( \omega \) is the angular velocity and \( R \) is the radius (1 m in this case).

### Step 3: Relate the frictional force to the centripetal force
The maximum static frictional force \( F_f \) that can act on the coin is given by:
\[ F_f = \mu N = \mu mg \]
Substituting \( N = mg \):
\[ F_f = \mu mg \]

### Step 4: Set the frictional force equal to the centripetal force
For the coin to not slip, the frictional force must be equal to the required centripetal force:
\[ \mu mg = m \omega^2 R \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \omega^2 R \]

### Step 5: Substitute the known values
Given:
- \( \mu = 0.5 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( R = 1 \, \text{m} \)

Substituting these values into the equation:
\[ 0.5 \times 10 = \omega^2 \times 1 \]
\[ 5 = \omega^2 \]
Thus, we find:
\[ \omega = \sqrt{5} \, \text{rad/s} \]

### Step 6: Use angular acceleration to find the time until slipping occurs
The disc starts from rest and has an angular acceleration \( \alpha = \frac{1}{\sqrt{2}} \, \text{rad/s}^2 \). We can use the kinematic equation for angular motion:
\[ \omega_f^2 = \omega_i^2 + 2 \alpha \theta \]
Where:
- \( \omega_f = \sqrt{5} \)
- \( \omega_i = 0 \)
- \( \alpha = \frac{1}{\sqrt{2}} \)
- \( \theta \) is the angle in radians.

Substituting the known values:
\[ 5 = 0 + 2 \left(\frac{1}{\sqrt{2}}\right) \theta \]
\[ 5 = \frac{2}{\sqrt{2}} \theta \]
\[ 5 = \sqrt{2} \theta \]
Thus, solving for \( \theta \):
\[ \theta = \frac{5}{\sqrt{2}} \, \text{radians} \]

### Step 7: Convert the angle from radians to revolutions
To find the number of revolutions \( n \):
\[ n = \frac{\theta}{2\pi} = \frac{5/\sqrt{2}}{2\pi} = \frac{5}{2\pi\sqrt{2}} \]

### Final Answer
The number of revolutions through which the disc turns before the coin slips is:
\[ n = \frac{5}{2\pi\sqrt{2}} \]