Question

An air-show pilot makes a vertical loop with a radius of curvature of 84 m.  Determine the normal force actin upon the 62.5 kg body at the bottom of the loop if the air speed is 62 m/s.

Answer 1

3472.62 N

Step-by-step explanation:

First, we need to make the free body diagram of the pilot at the bottom of the loop

Therefore, the net force is equal to

`F_(net)=F_n-mg=ma_c`

In a circular motion, the centripetal acceleration is equal to v²/r, so we will use the following equation

`\begin{gathered} F_n-mg=m(v^2)/(r) \\ \\ F_n=m(v^2)/(r)+mg \end{gathered}`

Where Fn is the normal force, m is the mass, v is the speed, r is the radius, and g is the gravity.

Replacing m = 62.5 kg, v = 62 m/s, r = 84 m, and g = 9.8 m/s², we get:

`\begin{gathered} F_n=\frac{(62.5\text{ kg\rparen\lparen62 m/s\rparen}^2}{84\text{ m}}+(62.5\text{ kg\rparen\lparen9.8 m/s}^2) \\ \\ F_n=2860.12\text{ N + 612.5 N} \\ F_n=3472.62\text{ N} \end{gathered}`

Therefore, the normal force is 3472.62 N