A cyclist moves around a circular path of radius `39.2sqrt(3)` metre with a speed of `19.6ms^(-1)` '. He must lean inwards at an angle `theta` with the vertical such that `tantheta` is

Options

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

Answer

To solve the problem, we need to determine the angle \( \theta \) at which the cyclist must lean while moving around a circular path. We will use the concepts of circular motion and the forces acting on the cyclist.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the circular path, \( r = 39.2\sqrt{3} \) m
- Speed of the cyclist, \( v = 19.6 \) m/s
- Acceleration due to gravity, \( g = 9.8 \) m/s²

2. **Understand the Forces Acting on the Cyclist:**
- The cyclist experiences a gravitational force \( mg \) acting downwards.
- The normal force \( N \) acts perpendicular to the surface of the circular path.
- The centripetal force required for circular motion is provided by the horizontal component of the normal force.

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

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

5. **Substitute the Known Values:**
- Substitute \( v = 19.6 \) m/s, \( r = 39.2\sqrt{3} \) m, and \( g = 9.8 \) m/s² into equation (3):
\[ \tan \theta = \frac{(19.6)^2}{(39.2\sqrt{3})(9.8)} \]

6. **Calculate the Right Side:**
- Calculate \( (19.6)^2 = 384.16 \).
- Calculate \( 39.2\sqrt{3} \approx 39.2 \times 1.732 = 68.0 \) (approximately).
- Now calculate \( 39.2\sqrt{3} \times 9.8 \approx 68.0 \times 9.8 \approx 666.4 \).
- Therefore:
\[ \tan \theta = \frac{384.16}{666.4} \approx 0.576 \]

7. **Simplify the Expression:**
- Recognizing that \( \tan \theta \approx \frac{1}{\sqrt{3}} \) implies that \( \theta \) is approximately \( 30^\circ \).

8. **Final Result:**
- Thus, we conclude that:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Final Answer:
The value of \( \tan \theta \) is \( \frac{1}{\sqrt{3}} \).

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