A cystlist riding a bicycle at a speed of `14sqrt(3) ` m/s takes a turn around a circular road of radius `20 sqrt(3)` m without skidding . What is his inclination to the vertical ?

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem of finding the cyclist's inclination to the vertical while riding around a circular road, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Speed of the cyclist, \( v = 14\sqrt{3} \, \text{m/s} \)
- Radius of the circular path, \( r = 20\sqrt{3} \, \text{m} \)
- Acceleration due to gravity, \( g \approx 9.8 \, \text{m/s}^2 \)

2. **Understand Forces Acting on the Cyclist**:
- The weight of the cyclist and the bicycle acts downward: \( mg \).
- The normal force \( N \) acts perpendicular to the surface of the bicycle.
- When the cyclist takes a turn, a centripetal force is required to keep moving in a circular path, which is provided by the horizontal component of the normal force.

3. **Set Up the Equations**:
- The vertical component of the normal force balances the weight:
\[ N \cos \theta = mg \quad \text{(1)} \]
- The horizontal component of the normal force provides the centripetal force:
\[ N \sin \theta = \frac{mv^2}{r} \quad \text{(2)} \]

4. **Divide Equation (2) by Equation (1)**:
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{v^2}{rg} \quad \text{(3)} \]

5. **Substitute the Known Values**:
- Calculate \( v^2 \):
\[ v^2 = (14\sqrt{3})^2 = 196 \times 3 = 588 \, \text{m}^2/\text{s}^2 \]
- Substitute \( v^2 \), \( r \), and \( g \) into equation (3):
\[ \tan \theta = \frac{588}{(20\sqrt{3})(9.8)} \]

6. **Calculate the Denominator**:
- First calculate \( 20\sqrt{3} \):
\[ 20\sqrt{3} \approx 34.64 \, \text{m} \]
- Now calculate \( 20\sqrt{3} \cdot 9.8 \):
\[ 34.64 \cdot 9.8 \approx 339.392 \, \text{m} \]

7. **Calculate \( \tan \theta \)**:
\[ \tan \theta = \frac{588}{339.392} \approx 1.73 \]

8. **Find the Angle \( \theta \)**:
- Recognizing that \( \tan 60^\circ = \sqrt{3} \approx 1.732 \), we can conclude:
\[ \theta \approx 60^\circ \]

9. **Conclusion**:
- The inclination of the cyclist to the vertical is \( 60^\circ \).

### Final Answer:
The inclination to the vertical is \( 60^\circ \).