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 solve the problem, we need to determine the minimum coefficient of friction (μ) required to prevent the coin from slipping off the rotating disc. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Angular speed (ω) = 2 rad/s
- Radius of the disc (R) = 50 cm = 0.5 m (convert to meters)
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the centrifugal force acting on the coin
The centrifugal force (Fc) acting on the coin can be calculated using the formula:
\[ F_c = m \cdot \omega^2 \cdot R \]
Substituting the values:
\[ F_c = m \cdot (2)^2 \cdot (0.5) \]
\[ F_c = m \cdot 4 \cdot 0.5 \]
\[ F_c = 2m \]

### Step 3: Identify the forces acting on the coin
The forces acting on the coin include:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The normal force (N) acting upwards: \( N = mg \)
- The frictional force (Ff) acting towards the center of the disc: \( F_f = \mu N \)

### Step 4: Set up the equilibrium condition
For the coin to not slip, the frictional force must be equal to the centrifugal force:
\[ F_f = F_c \]
Substituting the expressions for frictional force and centrifugal force:
\[ \mu N = F_c \]
Since \( N = mg \), we can substitute this into the equation:
\[ \mu (mg) = 2m \]

### Step 5: Simplify the equation
We can cancel the mass (m) from both sides (assuming m ≠ 0):
\[ \mu g = 2 \]
Now substituting the value of g:
\[ \mu (10) = 2 \]

### Step 6: Solve for the coefficient of friction (μ)
Now, we can solve for μ:
\[ \mu = \frac{2}{10} \]
\[ \mu = 0.2 \]

### Conclusion
The minimum coefficient of friction required to prevent the coin from slipping off the disc is:
\[ \mu = 0.2 \]

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