An aeroplane is flying at a velocity of `"900 km h"^(-1)` loops a vertical circular loop. If the maximum force pressing the pilot against the seat is five times his weight, what would be the diameter (in m) of the loop? `[g=10ms^(-2)]`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the velocity from km/h to m/s
The velocity of the airplane is given as 900 km/h. To convert this to meters per second, we use the conversion factor:
\[ 1 \text{ km/h} = \frac{5}{18} \text{ m/s} \]
Thus, we calculate:
\[ v = 900 \times \frac{5}{18} = 250 \text{ m/s} \]

### Step 2: Understand the forces acting on the pilot
When the airplane loops a vertical circular loop, the forces acting on the pilot at the bottom of the loop are:
- The gravitational force acting downwards: \( mg \)
- The normal force acting upwards: \( N \)

According to the problem, the maximum force pressing the pilot against the seat is five times his weight:
\[ N = 5mg \]

### Step 3: Apply Newton's second law
At the bottom of the loop, the net force acting on the pilot is the difference between the normal force and the weight:
\[ N - mg = \frac{mv^2}{r} \]
Substituting for \( N \):
\[ 5mg - mg = \frac{mv^2}{r} \]
This simplifies to:
\[ 4mg = \frac{mv^2}{r} \]

### Step 4: Cancel mass \( m \) from both sides
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ 4g = \frac{v^2}{r} \]

### Step 5: Solve for the radius \( r \)
Rearranging the equation gives:
\[ r = \frac{v^2}{4g} \]
Substituting the values \( v = 250 \text{ m/s} \) and \( g = 10 \text{ m/s}^2 \):
\[ r = \frac{(250)^2}{4 \times 10} = \frac{62500}{40} = 1562.5 \text{ m} \]

### Step 6: Calculate the diameter of the loop
The diameter \( D \) of the loop is twice the radius:
\[ D = 2r = 2 \times 1562.5 = 3125 \text{ m} \]

### Final Answer
The diameter of the loop is \( 3125 \text{ m} \).

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