A cyclist wants to loop a loop inside a death globe of radius 5 m. Calculate the least velocity the cyclist should have at the lowest point and calculate the height from which he should start ?

Explanation

To solve the problem, we need to calculate the least velocity the cyclist should have at the lowest point of the loop and the height from which he should start.

### Step 1: Calculate the minimum velocity at the lowest point

The minimum velocity \( v \) required at the lowest point of the loop can be derived from the condition that the centripetal force must be equal to the gravitational force acting on the cyclist at the top of the loop.

At the top of the loop, the gravitational force provides the necessary centripetal force. The formula for centripetal force is:

\[ F_c = \frac{mv^2}{r} \]

Where:
- \( m \) is the mass of the cyclist,
- \( v \) is the velocity at the top of the loop,
- \( r \) is the radius of the loop.

At the top of the loop, the gravitational force \( mg \) must equal the centripetal force:

\[ mg = \frac{mv^2}{r} \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)) gives:

\[ g = \frac{v^2}{r} \]

Rearranging this equation to find \( v \):

\[ v = \sqrt{gr} \]

Given that the radius \( r = 5 \) m and \( g \approx 9.8 \) m/s², we can substitute these values into the equation:

\[ v = \sqrt{9.8 \times 5} \]

Calculating this gives:

\[ v = \sqrt{49} = 7 \text{ m/s} \]

### Step 2: Calculate the height from which the cyclist should start

To find the height \( h \) from which the cyclist should start, we can use the principle of conservation of energy. The potential energy at the height \( h \) will be converted into kinetic energy at the lowest point of the loop.

The potential energy at height \( h \) is given by:

\[ PE = mgh \]

The kinetic energy at the lowest point is given by:

\[ KE = \frac{1}{2} mv^2 \]

Setting the potential energy equal to the kinetic energy:

\[ mgh = \frac{1}{2} mv^2 \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)) gives:

\[ gh = \frac{1}{2} v^2 \]

Rearranging to find \( h \):

\[ h = \frac{v^2}{2g} \]

Substituting \( v = 7 \) m/s and \( g = 9.8 \) m/s²:

\[ h = \frac{7^2}{2 \times 9.8} \]

Calculating this gives:

\[ h = \frac{49}{19.6} \approx 2.5 \text{ m} \]

### Final Results

- The least velocity the cyclist should have at the lowest point is **7 m/s**.
- The height from which he should start is approximately **2.5 m**.