You may have seen in a circus a motorcyclist driving in a vertical loop inside a 'death well' chamber with holes, so the spectators can watch from outside. Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop if the radius of the chamber is 25 m ?

Explanation

To solve the problem, we will break it down into two parts: explaining why the motorcyclist does not drop down when at the uppermost point of the loop, and calculating the minimum speed required to perform a vertical loop with a radius of 25 m.

### Step 1: Understanding Forces at the Uppermost Point
When the motorcyclist is at the top of the vertical loop, two forces act on him:
- The gravitational force (weight) acting downwards, which is equal to \( mg \) (where \( m \) is the mass of the motorcyclist and \( g \) is the acceleration due to gravity).
- The centripetal force required to keep the motorcyclist moving in a circular path, which is provided by the net force acting towards the center of the loop.

At the topmost point, the centripetal force is given by the formula:
\[ F_c = \frac{mv^2}{R} \]
where \( v \) is the velocity of the motorcyclist and \( R \) is the radius of the loop.

For the motorcyclist to stay in the loop without falling, the gravitational force must equal the required centripetal force:
\[ mg = \frac{mv^2}{R} \]
This simplifies to:
\[ g = \frac{v^2}{R} \]
This means that as long as the motorcyclist has enough speed, the gravitational force will provide the necessary centripetal force to keep him in the loop.

### Step 2: Finding the Minimum Speed
To find the minimum speed required to complete the loop, we can rearrange the equation derived above:
\[ v^2 = gR \]
To ensure the motorcyclist does not fall off, we need to consider the minimum speed at the top of the loop. The minimum speed can be derived as follows:
\[ v_{\text{min}} = \sqrt{gR} \]
However, to complete the entire loop, we need to consider that the motorcyclist should have enough speed at the bottom of the loop to maintain motion throughout. The formula for the minimum speed to complete the loop is given by:
\[ v_{\text{min}} = \sqrt{5gR} \]

Now, substituting \( R = 25 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \):
\[ v_{\text{min}} = \sqrt{5 \times 10 \times 25} \]
Calculating this gives:
\[ v_{\text{min}} = \sqrt{1250} \]
\[ v_{\text{min}} \approx 35.36 \, \text{m/s} \]

### Final Answers
1. The motorcyclist does not drop down at the uppermost point because the gravitational force provides the necessary centripetal force to keep him in circular motion.
2. The minimum speed required to perform a vertical loop with a radius of 25 m is approximately \( 35.36 \, \text{m/s} \).