A force F is applied at the centre of a disc of mass M. The minimum value of coefficient of friction of the surface for rolling is

Options

A. `(F)/(2Mg)`
B. `(F)/(3Mg)`
C. `(2F)/(5Mg)`
D. `(2F)/(7Mg)`

Answer

To solve the problem of finding the minimum value of the coefficient of friction (μ) for a disc of mass M rolling without slipping when a force F is applied at its center, we can follow these steps:

### Step 1: Understand the Forces Acting on the Disc
- A force \( F \) is applied at the center of the disc.
- The frictional force \( f \) acts in the opposite direction to the applied force to prevent slipping.
- The disc has a mass \( M \).

### Step 2: Write the Equation of Motion
From Newton's second law, we can write the equation for the net force acting on the disc:
\[ F - f = M \cdot a \quad \text{(1)} \]
where \( a \) is the linear acceleration of the center of mass of the disc.

### Step 3: Relate Linear Acceleration to Angular Acceleration
For pure rolling motion, the acceleration \( a \) of the center of mass and the angular acceleration \( \alpha \) are related by the equation:
\[ a = \alpha \cdot r \quad \text{(2)} \]
where \( r \) is the radius of the disc.

### Step 4: Write the Torque Equation
The torque \( \tau \) about the center of mass due to the frictional force is given by:
\[ \tau = f \cdot r = I \cdot \alpha \quad \text{(3)} \]
For a disc, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} M r^2 \]
Substituting this into the torque equation gives:
\[ f \cdot r = \frac{1}{2} M r^2 \cdot \alpha \]

### Step 5: Substitute Angular Acceleration
Using equation (2), we can express \( \alpha \) in terms of \( a \):
\[ f \cdot r = \frac{1}{2} M r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ f = \frac{1}{2} M a \quad \text{(4)} \]

### Step 6: Substitute Equation (4) into Equation (1)
Now we substitute equation (4) into equation (1):
\[ F - \frac{1}{2} M a = M a \]
Rearranging gives:
\[ F = M a + \frac{1}{2} M a = \frac{3}{2} M a \]
Thus, we can solve for \( a \):
\[ a = \frac{2F}{3M} \quad \text{(5)} \]

### Step 7: Find the Frictional Force
Substituting equation (5) back into equation (4) gives us the frictional force:
\[ f = \frac{1}{2} M \cdot \frac{2F}{3M} = \frac{F}{3} \quad \text{(6)} \]

### Step 8: Relate Frictional Force to Coefficient of Friction
The maximum static frictional force can be expressed as:
\[ f = \mu \cdot N \]
where \( N = Mg \) is the normal force. Thus:
\[ \frac{F}{3} = \mu \cdot Mg \]

### Step 9: Solve for the Coefficient of Friction
Rearranging gives:
\[ \mu = \frac{F}{3Mg} \]

### Final Answer
The minimum value of the coefficient of friction \( \mu \) for rolling without slipping is:
\[ \mu = \frac{F}{3Mg} \]

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