Pilot's speed at top of loop when weightless and apparent weight at bottom

[nissan300zxs]

Having trouble with a homework problem on Newtons Laws, would appreciate any help:
An airplane flies in a loop (a circular path in a vertical plane) of radius 150m. 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)At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point? b) At the bottom of the loop, the speed of the airplane is 280km/h. What is the apparent weight of the pilot at this point? His true weight is 700N.

Thanks in adavance for any assistance.

 

[Sirus]

This will be moved to homework section promptly, but anyways...

Please show us what you have done so far. Try beginning with a free-body diagram for each position. What can you deduce about the net force acting on the pilot if he feels weightless? You may find the following formulas useful:
$$F_{net}=ma$$
$$F_{centripetal}=\frac{mv^{2}}{r}$$

 

[wais]

Hi there,

I can definitely help you with this Newton's Laws problem. Let's break it down step by step.

a) At the top of the loop, the pilot feels weightless. This means that the net force acting on the pilot is zero, resulting in an apparent weight of zero. This is because at this point, the normal force (which is equal in magnitude but opposite in direction to the weight) is equal to the centripetal force (which is responsible for keeping the plane moving in a circular path). Using Newton's Second Law (F=ma), we can set up the following equation:

N - mg = mv^2/r

where N is the normal force, m is the mass of the pilot, g is the acceleration due to gravity, v is the speed of the airplane, and r is the radius of the loop. Since the pilot's apparent weight is zero, we can plug in N=0 and solve for v:

0 - mg = mv^2/r
0 - (m)(9.8 m/s^2) = (m)v^2/(150 m)
0 = v^2/150
v = √(0.0 m/s^2 * 150 m)
v = 0 m/s

This means that at the top of the loop, the speed of the airplane is 0 m/s.

b) At the bottom of the loop, the speed of the airplane is 280 km/h. We first need to convert this speed to m/s by dividing by 3.6:

280 km/h ÷ 3.6 = 77.78 m/s

Now, we can use the same equation as before to solve for the apparent weight of the pilot:

N - mg = mv^2/r
N - (700 N) = (m)(77.78 m/s)^2/(150 m)
N = (m)(77.78 m/s)^2/(150 m) + (700 N)
N = (m)(60.54 m/s^2) + (700 N)

Since we know that the pilot's true weight is 700 N, we can set N=700 N and solve for m:

700 N = (m)(60.54 m/s^2) + (700 N)
0 = (m)(60.54 m/s^2)
m = 0 kg

This means that the mass of the pilot is 0 kg