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. `(5)/(4pi)`
D. `(3)/(4pi)`

Answer

To solve the problem step by step, we need to analyze the forces acting on the coin and the motion of the disc.

### Step 1: Identify the forces acting on the coin
The coin experiences two main forces:
1. The gravitational force (weight) acting downwards: \( F_g = mg \)
2. The frictional force acting towards the center of the disc, which provides the necessary centripetal force for circular motion.

### Step 2: Write the expression for centripetal force
The centripetal force required to keep the coin moving in a circle of radius \( r \) is given by:
\[ F_c = \frac{mv^2}{r} \]
Where \( v \) is the linear velocity of the coin.

### Step 3: Relate linear velocity to angular velocity
The linear velocity \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = \omega r \]
Substituting this into the centripetal force equation gives:
\[ F_c = \frac{m(\omega r)^2}{r} = m\omega^2 r \]

### Step 4: Write the expression for frictional force
The maximum frictional force that can act on the coin before it slips is given by:
\[ F_f = \mu mg \]
Where \( \mu \) is the coefficient of friction.

### Step 5: Set up the equation for slipping condition
For the coin to remain on the disc, the centripetal force must not exceed the maximum frictional force:
\[ m\omega^2 r \leq \mu mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \omega^2 r \leq \mu g \]
Rearranging gives:
\[ \omega^2 \leq \frac{\mu g}{r} \]

### Step 6: Calculate the critical angular velocity
Substituting the given values \( \mu = 0.5 \), \( g = 9.8 \, \text{m/s}^2 \), and \( r = 1 \, \text{m} \):
\[ \omega^2 \leq \frac{0.5 \times 9.8}{1} = 4.9 \]
Taking the square root:
\[ \omega \leq \sqrt{4.9} \approx 2.2136 \, \text{rad/s} \]

### Step 7: Relate angular acceleration and angular displacement
The disc starts from rest, so the initial angular velocity \( \omega_0 = 0 \). The angular displacement \( \theta \) can be calculated using the equation:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Substituting \( \omega_0 = 0 \) and \( \alpha = \frac{1}{\sqrt{2}} \):
\[ \omega^2 = 2\left(\frac{1}{\sqrt{2}}\right)\theta \]
Setting \( \omega = 2.2136 \):
\[ (2.2136)^2 = 2\left(\frac{1}{\sqrt{2}}\right)\theta \]
Calculating \( (2.2136)^2 \):
\[ 4.9 = 2\left(\frac{1}{\sqrt{2}}\right)\theta \]
\[ \theta = \frac{4.9 \sqrt{2}}{2} = 4.9 \cdot \frac{1.4142}{2} \approx 3.465 \]

### Step 8: Calculate the number of revolutions
To find the number of revolutions \( N \), we convert the angular displacement from radians to revolutions:
\[ N = \frac{\theta}{2\pi} = \frac{3.465}{2\pi} \approx \frac{3.465}{6.2832} \approx 0.55 \]

### Final Answer
The number of revolutions through which the disc turns before the coin slips is approximately \( 0.55 \) revolutions.

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