91Ó°ÊÓ

The given data can be listed below as,

• The radius of the loop is,$R=150\text{″¾}$ .
• The weight of the pilot is, $F_{g,pilot}=700\text{ N}$.
• The speed of the airplane at the bottom is, $v_{bottom}=280\text{ k³¾/³ó}=280×\frac{5}{18}=77.8\text{″¾/²õ}$.

The acceleration of a body when the body moves in a circular path, its direction and velocity change continuously; it points towards the center of the circular path.

Part (a)

The expression for the centripetal acceleration can be expressed as,

$\begin{array}{c}{a}_r=g=\frac{v_{top}^2}{R}\\ v_{top}=\sqrt{gR}\end{array}$

Here$g$is the acceleration due to gravity,$R$is the radius of the loop.

Substitute $9.8{\text{″¾/²õ}}^{\text{2}}$for $g$, and $150\text{″¾}$for $R$in the above equation

$\begin{array}{c}{v}_{top}=\sqrt{9.8{\text{″¾/²õ}}^{\text{2}}×150\text{″¾}}\\ =38.3\text{″¾/²õ}\end{array}$

Hence, the required velocity of the airplane at the top is,$38.3\text{″¾/²õ}$ .

Part (b)

The expression for the apparent weight of the pilot at the bottom can be expressed as,

$F_{bottom}=\left(m_{pilot}{a}_r\right)+F_{g,pilot}$

$F_{bottom}=(\frac{F_{g,pilot}}{g}×\frac{v_{bottom}^2}{R})+F_{g,pilot}$

Here $F_{g,pilot}$is the weight of the pilot at the bottom of the loop,

Substitute $700\text{ N}$for $F_{g,pilot}$, $77.8\text{″¾/²õ}$for $v_{bottom}$, and $9.8{\text{″¾/²õ}}^{\text{2}}$for ,$g$and $150\text{″¾}$for $R$in the above equation.

$\begin{array}{c}{F}_{bottom}=(\frac{700\text{ N}}{9.8{\text{″¾/²õ}}^{\text{2}}}×\frac{{\left(77.8\text{″¾/²õ}\right)}^2}{150\text{″¾}})+700\text{ N}\\ =3582\text{ N}\end{array}$

Hence, the required apparent weight of the pilot at the bottom is, 3582 N