The least coefficient of friction for an inclined plane inclined at angle `alpha` with horizontal in order that a solid cylinder will roll down without slipping is

Options

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

Explanation

To find the least coefficient of friction for a solid cylinder rolling down an inclined plane without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Cylinder**:
- The gravitational force acting downwards: \( F_g = mg \).
- The component of gravitational force acting down the incline: \( F_{\text{down}} = mg \sin \alpha \).
- The normal force acting perpendicular to the incline: \( N = mg \cos \alpha \).
- The frictional force acting up the incline: \( F_f \).

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

3. **Relate Linear Acceleration to Angular Acceleration**:
- The frictional force causes a torque about the center of the cylinder:
\[ \tau = F_f \cdot r \]
- The moment of inertia \( I \) of a solid cylinder about its axis is given by:
\[ I = \frac{1}{2} m r^2 \]
- The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ a = r \alpha \]

4. **Write the Torque Equation**:
- Using the relation for torque:
\[ F_f \cdot r = I \alpha \]
- Substituting \( I \) and \( \alpha \):
\[ F_f \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
- Simplifying gives:
\[ F_f = \frac{1}{2} m a \]

5. **Substitute \( F_f \) in the Linear Motion Equation**:
- Replace \( F_f \) in the linear motion equation:
\[ mg \sin \alpha - \frac{1}{2} m a = ma \]
- Rearranging gives:
\[ mg \sin \alpha = \frac{3}{2} ma \]
- Thus, we can express \( a \):
\[ a = \frac{2g \sin \alpha}{3} \]

6. **Find the Frictional Force**:
- Now substitute \( a \) back to find \( F_f \):
\[ F_f = \frac{1}{2} m a = \frac{1}{2} m \cdot \frac{2g \sin \alpha}{3} = \frac{mg \sin \alpha}{3} \]

7. **Relate Frictional Force to Normal Force**:
- The maximum static friction force is given by:
\[ F_f \leq \mu_s N \]
- Substituting \( N = mg \cos \alpha \):
\[ \frac{mg \sin \alpha}{3} \leq \mu_s (mg \cos \alpha) \]
- Dividing through by \( mg \) (assuming \( m \neq 0 \)):
\[ \frac{\sin \alpha}{3} \leq \mu_s \cos \alpha \]

8. **Solve for the Coefficient of Friction**:
- Rearranging gives:
\[ \mu_s \geq \frac{\sin \alpha}{3 \cos \alpha} \]
- This can be simplified using the tangent function:
\[ \mu_s \geq \frac{1}{3} \tan \alpha \]

### Final Result:
The least coefficient of friction required for a solid cylinder to roll down an inclined plane without slipping is:
\[ \mu_s = \frac{1}{3} \tan \alpha \]