A solid cylinder rolls without slipping down a `30^@` slope. The minimum coefficient of friction needed to prevent slipping, will be

Options

A. 0.192
B. 0.18
C. 0.15
D. 0.2

Answer

To find the minimum coefficient of friction needed to prevent slipping for a solid cylinder rolling down a slope of \(30^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Cylinder:**
- The weight of the cylinder \(mg\) acts downward.
- The normal force \(N\) acts perpendicular to the slope.
- The frictional force \(f\) acts up the slope to prevent slipping.

2. **Resolve Weight into Components:**
- The component of the weight acting parallel to the slope is \(mg \sin \theta\).
- The component of the weight acting perpendicular to the slope is \(mg \cos \theta\).

3. **Apply Newton's Second Law:**
- For the translational motion along the slope:
\[ mg \sin \theta - f = ma \]
- For the rotational motion about the center of mass:
\[ f \cdot r = I \cdot \alpha \]
- For a solid cylinder, the moment of inertia \(I = \frac{1}{2} m r^2\) and the angular acceleration \(\alpha = \frac{a}{r}\).

4. **Substitute the Moment of Inertia:**
- Substitute \(I\) and \(\alpha\) into the rotational equation:
\[ f \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
- Simplifying gives:
\[ f = \frac{1}{2} ma \]

5. **Substitute \(f\) into the Translational Equation:**
- Replace \(f\) in the translational equation:
\[ mg \sin \theta - \frac{1}{2} ma = ma \]
- This simplifies to:
\[ mg \sin \theta = \frac{3}{2} ma \]

6. **Solve for Acceleration \(a\):**
- Rearranging gives:
\[ a = \frac{2g \sin \theta}{3} \]

7. **Relate Friction to Normal Force:**
- The normal force \(N\) is given by:
\[ N = mg \cos \theta \]
- The frictional force must satisfy:
\[ f \leq \mu N \]
- Substitute \(f = \frac{1}{2} ma\):
\[ \frac{1}{2} m \left(\frac{2g \sin \theta}{3}\right) \leq \mu mg \cos \theta \]

8. **Simplify the Inequality:**
- Cancel \(m\) from both sides:
\[ \frac{g \sin \theta}{3} \leq \mu g \cos \theta \]
- Dividing by \(g\) (assuming \(g \neq 0\)):
\[ \frac{\sin \theta}{3} \leq \mu \cos \theta \]

9. **Solve for the Coefficient of Friction \(\mu\):**
- Rearranging gives:
\[ \mu \geq \frac{\sin \theta}{3 \cos \theta} = \frac{1}{3} \tan \theta \]
- For \(\theta = 30^\circ\):
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
- Thus:
\[ \mu \geq \frac{1}{3\sqrt{3}} \]

### Final Answer:
The minimum coefficient of friction needed to prevent slipping is:
\[ \mu \geq \frac{1}{3\sqrt{3}} \approx 0.192 \]