A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a uniform angular speed 2 rad `s^(-1)`. The radius of the disc is 50 cm. Find the minimum coefficient of friction between disc and coin so that the coin does not slip. (Take `g = 10 m//s^(2)`)

Explanation

To find the minimum coefficient of friction between the disc and the coin so that the coin does not slip, we can follow these steps:

### Step 1: Identify the forces acting on the coin
When the disc rotates, the coin experiences a centripetal force directed towards the center of the disc. The forces acting on the coin are:
1. The gravitational force acting downwards: \( F_g = mg \)
2. The normal force acting upwards: \( N \)
3. The frictional force acting towards the center of the disc: \( F_f \)

### Step 2: Calculate the centripetal force required
The centripetal force (\( F_c \)) required to keep the coin moving in a circular path is given by:
\[ F_c = m \cdot a_c \]
where \( a_c \) is the centripetal acceleration. The centripetal acceleration can be calculated using:
\[ a_c = r \cdot \omega^2 \]
where:
- \( r = 0.5 \, \text{m} \) (the radius of the disc)
- \( \omega = 2 \, \text{rad/s} \)

Substituting the values:
\[ a_c = 0.5 \cdot (2)^2 = 0.5 \cdot 4 = 2 \, \text{m/s}^2 \]
Thus, the centripetal force is:
\[ F_c = m \cdot 2 \]

### Step 3: Relate frictional force to centripetal force
The frictional force (\( F_f \)) that prevents the coin from slipping is given by:
\[ F_f = \mu N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, the normal force is equal to the gravitational force:
\[ N = mg \]
Thus, we have:
\[ F_f = \mu mg \]

### Step 4: Set centripetal force equal to frictional force
For the coin to not slip, the frictional force must be equal to the centripetal force:
\[ \mu mg = m \cdot 2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = 2 \]

### Step 5: Solve for the coefficient of friction
Now we can solve for \( \mu \):
\[ \mu = \frac{2}{g} \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ \mu = \frac{2}{10} = 0.2 \]

### Final Answer
The minimum coefficient of friction between the disc and the coin so that the coin does not slip is \( \mu = 0.2 \).

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