Chapter 5, Problem 12P

(a)

To determine

The minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed $\text{6}\text{.0g's}$ .

Answer to Problem 12P

Solution:

The minimum radius of the loop so that centripetal acceleration at the lowest point does not exceed $\text{6}\text{.0}\text{g's}$

  $\text{=}\text{925}\text{.9m}$ .

Explanation of Solution

Given:

Jet speed, $v=\text{840}\text{km/h}$

Centripetal acceleration, $a=\text{6}\text{.0}\text{g}$

Mass of pilot, $m\text{=}\text{78}\text{kg}$

Formula Used:

Centripetal acceleration of circular motion

  $a_c=\frac{v^2}{r}$

Where,

v : speed of the body

r : radius of the circular path

$a_c$: Centripetal acceleration

Calculation:

  $\begin{array}{l}{a}_c\le 6.0g\\ \frac{v^2}{r}\le 6.0g\\ v=\frac{840\text{km}}{1\text{h}}\\ =\frac{840\times 1000}{60\times 60}\text{m/s}\\ =233.33\text{m/s}\\ \frac{v^2}{r}\le 6.0g\\ =\frac{{\left(233.33\right)}^2}{r}\le 6.0\times 9.80\\ r\ge \frac{{\left(233.33\right)}^2}{6.0\times 9.80}\\ r\ge 925.9\text{m}\end{array}$

Conclusion:

Minimum radius of the circle, so that centripetal acceleration does not exceed $\text{6}\text{.0}\text{g is equal to 925}\text{.9}\text{m}$ .

(b)

To determine

The effective weight of $\text{78-kg}$ pilot’s (the force with which the seat pushes up on him) at the bottom of the circle.

Answer to Problem 12P

Solution:

Effective weight of pilot at the bottom of the circle

  $\begin{array}{l}\text{=}\text{7}mg\\ \text{=}\text{5350}\text{.8}\text{N}\end{array}$

Explanation of Solution

From free body diagram,

  

Newton’s second law at the bottom of the loop

  $\begin{array}{l}{N}_b-mg=m{a}_c\\ N_b-mg=\frac{m{v}^2}{r}\\ N_b=mg+\frac{m{v}^2}{r}\\ N_b=m\left(g+\frac{v^2}{r}\right)\end{array}$

Centripetal acceleration for fixed radius and speed is the same $\frac{v^2}{r}=6g$

  $\begin{array}{l}{N}_b=m\left(g+6g\right)\\ =7mg\\ N_b=7mg\\ =7\times 78g\\ =546g\end{array}$

Conclusion:

Effective weight of pilot at the bottom of the circle $\text{=546 g}$

  $=(546\times 9.8)\text{N}=5350.8\text{N}$

(c)

To determine

At the top of the circle (assume the same speed).

Answer to Problem 12P

Solution:

Effective weight of pilot at the top of the circle

  $\begin{array}{l}\text{= 5}m\text{g}\\ \text{=}\text{3822}\text{N}\end{array}$

Explanation of Solution

From Free body diagram.

  

Newton’s second law at the top of the loop

  $\begin{array}{l}mg-N_t=m{a}_c\\ N_t=mg-m{a}_c\\ =m(g-a_c)\\ =m\left(g-6g\right)\left(a_c=6g\right)\\ N_t=-5mg\end{array}$

Negative normal force means it acts in downward direction.

So effective weight is present at top of the loop.

The = Normal force exerted by shoat belt in down direction

  $\begin{array}{l}=5mg\\ =5\times 78\times 9.8\\ =3822\text{N}\end{array}$

Conclusion:

Hence, effective weight at the top of the loop $\text{=}\text{3822}\text{N}$ .