A cyclist goes around a circular track of circumference 410 m in 20 s. Find the angle that his cycle makes with the vertical.

Explanation

To solve the problem of finding the angle that the cyclist's cycle makes with the vertical while going around a circular track, we can follow these steps:

### Step 1: Calculate the radius of the circular track
The circumference (C) of the circular track is given as 410 m. We can use the formula for the circumference of a circle:

\[ C = 2\pi r \]

Where \( r \) is the radius. Rearranging the formula to find \( r \):

\[ r = \frac{C}{2\pi} = \frac{410}{2\pi} \]

Calculating this gives:

\[ r \approx \frac{410}{6.2832} \approx 65.29 \text{ m} \]

### Step 2: Calculate the speed of the cyclist
The time (t) taken to complete one round of the track is given as 20 seconds. The speed (v) can be calculated using the formula:

\[ v = \frac{C}{t} = \frac{410}{20} \]

Calculating this gives:

\[ v = 20.5 \text{ m/s} \]

### Step 3: Use the formula for the angle with the vertical
In circular motion, the angle \( \theta \) that the cyclist makes with the vertical can be found using the relationship:

\[ \tan \theta = \frac{v^2}{rg} \]

Where:
- \( v \) is the speed of the cyclist (20.5 m/s),
- \( r \) is the radius (65.29 m),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the formula:

\[ \tan \theta = \frac{(20.5)^2}{65.29 \times 9.8} \]

Calculating \( v^2 \):

\[ v^2 = 20.5^2 = 420.25 \]

Calculating \( rg \):

\[ rg = 65.29 \times 9.8 \approx 639.842 \]

Now substituting these values into the equation for \( \tan \theta \):

\[ \tan \theta = \frac{420.25}{639.842} \approx 0.6568 \]

### Step 4: Calculate the angle \( \theta \)
To find \( \theta \), we take the arctangent (inverse tangent) of \( 0.6568 \):

\[ \theta = \tan^{-1}(0.6568) \approx 33.3^\circ \]

### Final Answer
The angle that the cyclist's cycle makes with the vertical is approximately \( 33.3^\circ \).

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