Question

An $85,000$kg stunt plane performs a loop-the-loop, flying in a $260-$m-diameter vertical circle. At the point where the plane is flying straight down, its speed is $55m/s$and it is speeding up at a rate of $12m/s$per second.

a. What is the magnitude of the net force on the plane? You can neglect air resistance.

b. What angle does the net force make with the horizontal? Let an angle above horizontal be positive and an angle below horizontal be negative.

Short answer

a). The magnitude of the net force on the plane is $2.23\times {10}^{6}N$.

b). The angle is $27.{28}^o$below the horizontal.

Step-by-step solution

Given Information (Part a)

An $85,000kg$stunt plane performs a loop-the-loop, flying in a $260-m$-diameter vertical circle.

Explanation (Part a)

Let's find the radius of the vertical circle:

$r=\frac{d}{2}$

Where, $r$is the radius

$d$is the diameter

$=\frac{260m}{2}$

$=130m$

The centripetal force acting on the plane is,

$F_1=\frac{m{v}^2}{r}$

Where, $m$is the mass of the plane

$v$is the velocity

$F_1=\frac{\left(85000kg\right){(55m/s)}^2}{\left(130m\right)}$

$=1.98\times {10}^{6}N$

Step 3:The magnitude of the net force(Part a)

The force acting on the plane due to radial acceleration is:

$F_2=ma$

$F_2=(85000kg)(12m/s^2)$

$=1.02\times {10}^{6}N$

The magnitude of the net force on the plane is :

$F=\sqrt{{F_1}^2+{F_2}^2}$

$F=\sqrt{{(1.98\times {10}^{6}N)}^2+{(1.02\times {10}^{6}N)}^2}$

$=2.23\times {10}^{6}N$

Final Answer (Part a)  

The magnitude of the net force on the plane is $2.23\times {10}^6\mathrm{N}$.

Given Information (Part b)

An $85,000kg$stunt plane performs a loop-the-loop, flying in a $260-m$-diameter vertical circle.

Explanation (Part b)

The angle is below the horizontal.

Therefore, the direction of force is

$\theta ={\tan }^{-1}\left(\frac{-F_2}{F_1}\right)$

$\theta ={\tan }^{-1}\left(\frac{-F_2}{F_1}\right)={\tan }^{-1}\left(\frac{-1.020\times {10}^6\mathrm{N}}{1.98\times {10}^6\mathrm{N}}\right)=-2.28°$

The angle is $27.28°$below the horizontal as it is negative.

Final Answer (part b)

The angle is $27.{28}^o$below the horizontal.