A cyclist is moving on a smooth horizontal curved path of radius of curvature 10m.with a speed 10 `ms^(-1)`. Then his angle of leaning is

Options

A. A) `15^(@)`
B. B) `30^(@)`
C. C) `45^(@)`
D. D) `60^(@)`

Answer

To find the angle of leaning of the cyclist moving on a curved path, we can use the formula for the angle of leaning, which is given by:

\[ \theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \]

Where:
- \( \theta \) is the angle of leaning,
- \( v \) is the speed of the cyclist,
- \( r \) is the radius of curvature,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \), but we can use \( 10 \, \text{m/s}^2 \) for simplicity).

### Step-by-step solution:

1. **Identify the given values:**
- Speed of the cyclist, \( v = 10 \, \text{m/s} \)
- Radius of curvature, \( r = 10 \, \text{m} \)
- Acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \)

2. **Substitute the values into the formula:**
\[ \theta = \tan^{-1}\left(\frac{v^2}{rg}\right) = \tan^{-1}\left(\frac{10^2}{10 \cdot 10}\right) \]

3. **Calculate \( v^2 \):**
\[ v^2 = 10^2 = 100 \]

4. **Calculate \( rg \):**
\[ rg = 10 \cdot 10 = 100 \]

5. **Substitute these values back into the equation:**
\[ \theta = \tan^{-1}\left(\frac{100}{100}\right) = \tan^{-1}(1) \]

6. **Find the angle whose tangent is 1:**
\[ \theta = 45^\circ \]

### Final Answer:
The angle of leaning of the cyclist is \( 45^\circ \).

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