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 for rolling when a force \( F \) is applied at the center of a disc of mass \( M \), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Forces Acting on the Disc**:
- A force \( F \) is applied at the center of the disc.
- The frictional force \( f \) acts at the point of contact with the surface to prevent slipping.
- The normal force \( N \) acts vertically upward, equal to the weight of the disc, \( N = Mg \).

2. **Set Up the Equations of Motion**:
- For the disc to roll without slipping, the condition \( a = R\alpha \) must hold, where \( a \) is the linear acceleration of the center of mass, \( R \) is the radius of the disc, and \( \alpha \) is the angular acceleration.
- The net force acting on the disc in the horizontal direction is given by:
\[ F - f = Ma \]

3. **Torque About the Center of Mass**:
- The torque \( \tau \) about the center of mass due to the frictional force is:
\[ \tau = fR \]
- According to Newton's second law for rotation, this torque is also equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \):
\[ fR = I\alpha \]

4. **Moment of Inertia of the Disc**:
- The moment of inertia \( I \) of a disc about its center is given by:
\[ I = \frac{1}{2}MR^2 \]

5. **Relate Angular Acceleration to Linear Acceleration**:
- Using the relationship \( \alpha = \frac{a}{R} \), we can substitute \( \alpha \) into the torque equation:
\[ fR = \left(\frac{1}{2}MR^2\right)\left(\frac{a}{R}\right) \]
- Simplifying this gives:
\[ fR = \frac{1}{2}MaR \]
- Dividing both sides by \( R \):
\[ f = \frac{1}{2}Ma \]

6. **Substituting for Acceleration**:
- From the linear motion equation \( F - f = Ma \), substitute \( f \):
\[ F - \frac{1}{2}Ma = Ma \]
- Rearranging gives:
\[ F = \frac{3}{2}Ma \]
- Therefore, solving for \( a \):
\[ a = \frac{2F}{3M} \]

7. **Substituting Back to Find Friction**:
- Now substitute \( a \) back into the friction equation:
\[ f = \frac{1}{2}M\left(\frac{2F}{3M}\right) = \frac{F}{3} \]

8. **Finding the Coefficient of Friction**:
- The frictional force must also satisfy the condition \( f = \mu N \), where \( N = Mg \):
\[ \mu Mg = \frac{F}{3} \]
- Thus, the coefficient of friction \( \mu \) is:
\[ \mu = \frac{F}{3Mg} \]

### Final Answer
The minimum value of the coefficient of friction for rolling is:
\[ \mu = \frac{F}{3Mg} \]