The minimum radius of a plane's circular path

[shootingrubbe]

Homework Statement

An 82 kg pilot flying a stunt airplane pulls out of a dive at a constant speed of 540 km/h.
a) What is the minimum radius of the plane's circular path if the pilot's acceleration at the lowest point is not to exceed 7.0g.

Homework Equations

Fg = mg
Fnet = mv^2/r
Fnet = ma

The Attempt at a Solution

At the lowest point I said that there were only 2 forces acting on the plane: Fg and I said Fn for the upward lift on the plane.

m=82 kg
v=540 km/h=150 m/s
a=7.0g

Fnet = mv^2/r=ma
Fnet = Fn - Fg = ma
Fnet = (82 kg)(7 x 9.8 N/kg) - (82 kg)(9.8 N/kg)
= 4821.6 N = mv^2/r
4821.6 N = [(82 kg)(150 m/s)^2]/r
r=[(82 kg)(150 m/s)^2]/4821.6 N
= 382.65 m
= 3.8 x 10^2 m

The answer is actually 3.3 x 10^2 m... The book is sometimes wrong though.

What did I do wrong?

Thanks.

 

[Vykan12]

I'm not sure what you did (your calculations are a bit messy), but here are the two equations I derived:

$$n= m(g+a_{y}) = 6435.36 N$$

$$r = \frac{mv^{2}}{n-mg} = \frac{(82)(150^{2})}{6435.36-(82)(9.81)} = 3.27 \cdot 10^{2} \, m$$

 

[shootingrubbe]

I'm not sure what you did (your calculations are a bit messy), but here are the two equations I derived:

$$n= m(g+a_{y}) = 6435.36 N$$

$$r = \frac{mv^{2}}{n-mg} = \frac{(82)(150^{2})}{6435.36-(82)(9.81)} = 3.27 \cdot 10^{2} \, m$$

That's essentially what I did, but I thought that the net force was Fn - Fg... How come it is Fn + Fg?

Attached is what I thought the FBD should look like.

 

[Vykan12]

$$F_{net} = n - mg = ma \iff n = mg + ma = m(g+a)$$

 

[shootingrubbe]

$$F_{net} = n - mg = ma \iff n = mg + ma = m(g+a)$$

What's n? I think we use different symbols where I take this course...

 

[Vykan12]

n is the normal force of the pilot.

 

[shootingrubbe]

n is the normal force of the pilot.

Ok, that's what I thought... Thanks.