Question

A jet pilot takes his aircraft in a vertical loop, as shown.

A jet pilot takes his aircraft in a vertical loop, as shown. 1.If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 7.0 g's? Calculate the 69-kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle, and at the top of the circle (assume the same speed)?

1. If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 7.0 g's?
2. Calculate the 69-kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle, and at the top of the circle (assume the same speed)?

Answer 1

Concepts and reason

The concepts used to solve this problem are centripetal force and Newton’s second law of motion.

Use the expressions for centripetal force and Newton’s second law of motion to determine the radius of the path.

Use the expression for normal force at the bottom of the circular path and at the top to find the effective weight of the pilot at the bottom and top of the circle respectively.

Fundamentals

The centripetal force is the one acting on a body moving in a circular path and the direction of the force is towards the center of the curvature of the path.

The expression for centripetal force is,

${F_{\rm{c}}} = \frac{{m{v^2}}}{r}$

Here,${F_{\rm{c}}}$is the centripetal force,$m$ is the mass of the body,$v$ is its velocity and $r$ is the radius of the circular path.

According to Newton’s second law of motion, the rate of change of momentum of a body is directly proportional to the unbalanced external force acting on it and the expression for this external force is,

$F = \frac{{dP}}{{dt}}$

Here, $F$ is the external force acting on the body, $P$ is its momentum and $t$ is the time.

Another expression for Newton’s second law of motion is,

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

(1)

The expression for centripetal force acting on the jet is,

${F_{\rm{c}}} = \frac{{m{v^2}}}{r}$

According to Newton’s second law, the net force acting on the jet is,

$F = ma$

Equating the above two expressions, the equation for radius of the path is,

$\begin{array}{c}\\\frac{{m{v^2}}}{r} = ma\\\\r = \frac{{{v^2}}}{a}\\\end{array}$

Substitute $1140\,{\mathop{\rm km}\nolimits} /h$ for $v$ and $7(9.8\,{\mathop{\rm m}\nolimits} /{s^2})$ for $a$.

$\begin{array}{c}\\r = \frac{{{{\left( {(1140\,{\rm{km/hr}})\left( {\frac{5}{{18}}{\rm{m/s}}} \right)\,} \right)}^2}}}{{(7)(9.8{\mathop{\rm m}\nolimits} /{s^2})}}\\\\ = 1.462 \times {10^3}{\mathop{\rm m}\nolimits} \\\end{array}$

(2.1)

The expression for effective weight of the pilot at the bottom of the circle is,

$N = mg + \frac{{m{v^2}}}{r}$

Substitute $69\,kg$for $m$,$9.8\,m/{s^2}$for $g$,$1140\,{\mathop{\rm km}\nolimits} /hr$ for $v$and $1.46 \times {10^3}\,{\mathop{\rm m}\nolimits}$for $r$.

$\begin{array}{c}\\N = (69\,{\mathop{\rm kg}\nolimits} )(9.8\,{\mathop{\rm m}\nolimits} /{s^2}) + \frac{{(69\,{\mathop{\rm kg}\nolimits} ){{\left( {(1140\,{\rm{km/hr}})\left( {\frac{5}{{18}}} \right)\,{\rm{m/s}}} \right)}^2}}}{{1.462 \times {{10}^3}\,{\mathop{\rm m}\nolimits} }}\\\\ = 5408.87\,{\mathop{\rm N}\nolimits} \\\end{array}$

(2.2)

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

$N = mg - \frac{{m{v^2}}}{r}$

Substitute $69\,kg$for $m$,$9.8\,m/{s^2}$for $g$,$1140\,{\mathop{\rm km}\nolimits} /hr$ for $v$and $1.462 \times {10^3}\,{\mathop{\rm m}\nolimits}$for $r$.

$\begin{array}{c}\\N = (69\,{\mathop{\rm kg}\nolimits} )(9.8\,{\mathop{\rm m}\nolimits} /{s^2}) - \frac{{(69\,{\mathop{\rm kg}\nolimits} ){{\left( {(1140\,{\rm{km/hr}})\left( {\frac{5}{{18}}} \right)\,{\rm{m/s}}} \right)}^2}}}{{1.462 \times {{10}^3}\,{\mathop{\rm m}\nolimits} }}\\\\ = - 4056.47\,{\mathop{\rm N}\nolimits} \\\end{array}$

The negative sign indicates the direction of force. So, weight at the top is,

$N = 4056.47{\mathop{\rm N}\nolimits}$

Ans: Part 1

The minimum radius of the circle is ${\bf{1}}{\bf{.462 \times 1}}{{\bf{0}}^{\bf{3}}}\,{\bf{m}}$.

Part 2.1

The pilot's effective weight at the bottom of the circle is ${\bf{5408}}{\bf{.87}}\,{\bf{N}}$.

Part 2.2

The pilot's effective weight at the top of the circle is ${\bf{4056}}{\bf{.47}}\,{\bf{N}}$.