An aircraft loops the loop of radius R = 500 m with a constant velocity ` v = 360km//h`. The weight of the flyer of mass m = 70 kg in the lower, upper and middle points of the loop will respectively be-

Options

A. `2.1 "kN", 0.7 kN, 1.5 kN`
B. 0.7 kN, 1.5 kN, 2.1 kN
C. 1.5 kN,2.1 kN, 0.7 kN
D. None of these

Answer

To solve the problem, we need to analyze the forces acting on the flyer at three different points in the loop: the lower point, the upper point, and the middle point.

### Given Data:
- Radius of the loop, \( R = 500 \, \text{m} \)
- Velocity of the aircraft, \( v = 360 \, \text{km/h} \)
- Mass of the flyer, \( m = 70 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 1: Convert the velocity from km/h to m/s
To convert the velocity from kilometers per hour to meters per second, we use the conversion factor \( \frac{5}{18} \).

\[ v = 360 \, \text{km/h} \times \frac{5}{18} = 100 \, \text{m/s} \]

### Step 2: Calculate the forces at the lower point of the loop
At the lower point of the loop, the forces acting on the flyer are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.

Using Newton's second law, we can write the equation for the forces:

\[ N - mg = \frac{mv^2}{R} \]

Rearranging gives:

\[ N = mg + \frac{mv^2}{R} \]

Substituting the values:

\[ N = 70 \times 10 + \frac{70 \times (100)^2}{500} \]

Calculating:

\[ N = 700 + \frac{70 \times 10000}{500} = 700 + 1400 = 2100 \, \text{N} \]

Converting to kilonewtons:

\[ N = 2.1 \, \text{kN} \]

### Step 3: Calculate the forces at the upper point of the loop
At the upper point of the loop, the forces acting on the flyer are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) also acting downward.

The equation for the forces is:

\[ N + mg = \frac{mv^2}{R} \]

Rearranging gives:

\[ N = \frac{mv^2}{R} - mg \]

Substituting the values:

\[ N = \frac{70 \times (100)^2}{500} - 70 \times 10 \]

Calculating:

\[ N = 1400 - 700 = 700 \, \text{N} \]

Converting to kilonewtons:

\[ N = 0.7 \, \text{kN} \]

### Step 4: Calculate the forces at the middle point of the loop
At the middle point of the loop, the forces acting on the flyer are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.

At this point, the normal force is equal to the centripetal force required to keep the flyer moving in a circular path:

\[ N = \frac{mv^2}{R} \]

Substituting the values:

\[ N = \frac{70 \times (100)^2}{500} \]

Calculating:

\[ N = 1400 \, \text{N} \]

Converting to kilonewtons:

\[ N = 1.4 \, \text{kN} \]

### Summary of Forces:
- At the lower point: \( N = 2.1 \, \text{kN} \)
- At the upper point: \( N = 0.7 \, \text{kN} \)
- At the middle point: \( N = 1.4 \, \text{kN} \)

### Final Answer:
The weights of the flyer at the lower, upper, and middle points of the loop will respectively be:
- Lower point: \( 2.1 \, \text{kN} \)
- Upper point: \( 0.7 \, \text{kN} \)
- Middle point: \( 1.4 \, \text{kN} \)