Chapter 5, Problem 73GP

(a)

To determine

The minimum radius of the circle.

Answer to Problem 73GP

The minimum radius of the vertical circle is $2367.41\text{m}$ .

Explanation of Solution

Given info:

The speed of the jet aircraft is, $v=1300\text{km}/\text{h}=\left(1300\text{km}/\text{h}\right)\left(\frac{0.287\text{m}/\text{s}}{1\text{km}/\text{h}}\right)=373.1\text{m}/\text{s}$ .

The acceleration of the lowest point is, $a=6g=6\left(9.8\text{m}/{\text{s}}^2\right)=58.8\text{m}/{\text{s}}^2$

Formula Used:

The expression to calculate the minimum radius of the vertical circle is,

  $r=\frac{v^2}{a}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}r=\frac{{\left(373.1\text{m}/\text{s}\right)}^2}{58.8\text{m}/{\text{s}}^2}\\ =2367.41\text{m}\end{array}$

Conclusion:

Thus, the minimum radius of the vertical circle is $2367.41\text{m}$ .

(b)

To determine

The effective weight of the pilot at the bottom of the circle.

Answer to Problem 73GP

The effective weight of the pilot at the bottom of the circle is $5350.8\text{N}$ .

Explanation of Solution

Given info:

The mass of the pilot is, $m=78\text{kg}$ .

Formula Used:

The expression to calculate theeffective weight of the pilot at the bottom of the circle is,

  $W_{eb}=m\left(\frac{v^2}{r}+g\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{W}_{eb}=\left(78\text{kg}\right)\left(\frac{{\left(373.1\text{m}/\text{s}\right)}^2}{2367.41\text{m}}+9.8\text{m}/{\text{s}}^2\right)\\ =5350.8\text{N}\end{array}$

Conclusion:

Thus, the effective weight of the pilot at the bottom of the circle is $5350.8\text{N}$ .

(c)

To determine

The effective weight of the pilot at the top of the circle.

Answer to Problem 73GP

The effective weight of the pilot at the top of the circle is $-3636.12\text{N}$ .

Explanation of Solution

The expression to calculate the effective weight of the pilot at the top of the circle is,

  $W_{et}=m\left(g-\frac{v^2}{r}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{W}_{et}=\left(78\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2-\frac{{\left(373.1\text{m}/\text{s}\right)}^2}{2367.41\text{m}}\right)\\ =-3636.12\text{N}\end{array}$

Conclusion:

Thus, the effective weight of the pilot at the top of the circle is $-3636.12\text{N}$ .