Q5-55E - University Physics with Modern Physics

Question

An airplane flies in a loop (a circular path in a vertical plane) of radius 150 m. The pilot’s head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom. 

(a) What is the speed of the airplane at the top of the loop, where the pilot feels weightless? 

(b) What is the apparent weight of the pilot at the bottom of the loop, where the speed of the airplane is $280\text{\hspace{0.17em}km/h}$? His true weight is 700 N.

Step-by-Step Solution

Answer

(a) The speed of the airplane at the top of the loop is, $38.3\text{\hspace{0.17em}m/s}$.

(b) The apparent weight of the pilot at the bottom of the loop is, 3582 N.

Step 1: Identification of given data

The given data can be listed below as,

• The radius of the loop is,$R=150\text{\hspace{0.17em}m}$ .
• The weight of the pilot is, $F_{g,pilot}=700\text{\hspace{0.17em}N}$.
• The speed of the airplane at the bottom is, $v_{bottom}=280\text{\hspace{0.17em}km/h}=280\times \frac{5}{18}=77.8\text{\hspace{0.17em}m/s}$. 

Step 2: Significance of centripetal acceleration

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.

Step 3: Determination of the speed of the airplane at the top of the loop

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 $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for $g$ , and $150\text{\hspace{0.17em}m}$ for $R$ in the above equation

$\begin{array}{c}{v}_{top}=\sqrt{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\times 150\text{\hspace{0.17em}m}}\\ =38.3\text{\hspace{0.17em}m/s}\end{array}$

 

Hence, the required velocity of the airplane at the top is,$38.3\text{\hspace{0.17em}m/s}$ .

Step 4: Determination of the apparent weight of the pilot at the bottom of the loop

Part (b)

 

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

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

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

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

 

Substitute $700\text{\hspace{0.17em}N}$ for $F_{g,pilot}$ , $77.8\text{\hspace{0.17em}m/s}$ for $v_{bottom}$ , and $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for ,$g$ and $150\text{\hspace{0.17em}m}$ for $R$ in the above equation.

 $\begin{array}{c}{F}_{bottom}=(\frac{700\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}\times \frac{{\left(77.8\text{\hspace{0.17em}m/s}\right)}^2}{150\text{\hspace{0.17em}m}})+700\text{\hspace{0.17em}N}\\ =3582\text{\hspace{0.17em}N}\end{array}$

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