A solid cylinder is rolling down on an inclined plane of angle `theta`. The minimum value of the coefficient of friction between the plane and the cylinder to allow pure rolling

Options

A. `(1)/(3)tantheta`
B. `tantheta`
C. `(2)/(3)tantheta`
D. `3tantheta`

Answer

To find the minimum value of the coefficient of friction (\( \mu \)) required for a solid cylinder to roll down an inclined plane without slipping, we can follow these steps:

### Step-by-Step Solution

1. **Identify Forces Acting on the Cylinder**:
- The weight of the cylinder (\( mg \)) acts vertically downward.
- The component of the weight acting down the incline is \( mg \sin \theta \).
- The normal force (\( N \)) acts perpendicular to the inclined plane, given by \( N = mg \cos \theta \).
- The frictional force (\( F \)) acts up the incline to prevent slipping.

2. **Apply Newton's Second Law**:
- For translational motion along the incline:
\[ mg \sin \theta - F = ma \]
- Here, \( a \) is the linear acceleration of the center of mass of the cylinder.

3. **Torque and Rotational Motion**:
- The frictional force provides the torque that causes the cylinder to rotate:
\[ \tau = I \alpha \]
- For a solid cylinder, the moment of inertia (\( I \)) about its center is \( \frac{1}{2} m r^2 \).
- The angular acceleration (\( \alpha \)) is related to the linear acceleration (\( a \)) by \( a = r \alpha \), so \( \alpha = \frac{a}{r} \).

4. **Express Torque in Terms of Friction**:
- The torque due to friction is:
\[ \tau = F \cdot r \]
- Setting the two expressions for torque equal gives:
\[ F \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
- Simplifying this leads to:
\[ F = \frac{1}{2} ma \]

5. **Substitute for Friction in the Translational Equation**:
- Substitute \( F = \mu N = \mu mg \cos \theta \) into the translational motion equation:
\[ mg \sin \theta - \mu mg \cos \theta = ma \]
- Rearranging gives:
\[ mg \sin \theta = ma + \mu mg \cos \theta \]

6. **Combine Equations**:
- Substitute \( a = \frac{1}{2} F \) into the equation:
\[ mg \sin \theta = \frac{1}{2} \mu mg \cos \theta + \mu mg \cos \theta \]
- This simplifies to:
\[ mg \sin \theta = \frac{3}{2} \mu mg \cos \theta \]

7. **Solve for \( \mu \)**:
- Cancel \( mg \) from both sides:
\[ \sin \theta = \frac{3}{2} \mu \cos \theta \]
- Rearranging gives:
\[ \mu = \frac{2}{3} \frac{\sin \theta}{\cos \theta} = \frac{2}{3} \tan \theta \]

8. **Final Result**:
- Therefore, the minimum value of the coefficient of friction required for pure rolling is:
\[ \mu = \frac{2}{3} \tan \theta \]