Circular loop problem: finding the radius of a small circle

1. Sep 15, 2008

Benzoate

1. The problem statement, all variables and given/known data

In an air show, a pilot is to execute a circular loop at the speed of sound(340 m/s) . The pilot may black out if his acceleration exceeds 8g. Find the radius of the smallest circle he can use. [Take g=10 m s^-2

2. Relevant equations

possible equations: u^2 >= 2*M*G/a
m*dv/dt=-m*M*G/r^2
m*v^2/r=-m*M*G/r^2
3. The attempt at a solution

u=340 m/s
a=dv/dt=>=80 m s^-2
M=6.00e24 kg
G=6.67e-11

acceleration=dv/dt=> dv/dt=-M*G/r^2

option one
r=sqrt(M*G/(dv/dt)=7.07e12 meters
option 2
a=2*M*G/u^2, a being minimal radius and not the acceleration

a= 28284271.25 meters

what did I do wrong? should I have used polar coordinates since the problem states that the plane is going around a loop?

Perhaps I should write:

m*dv/dt=F(z) , F(z) representing the sum of all forces. the only two forces acting on the object is the gravitational force and the centripetal force ; so my equation looks like:

m*dv/dt=m*v^2/r-m*g

dv/dt=dv/dr*dr/dt=dv/dr*v. Now I can integrate in terms of the velocity vector and radius vector.

dv/dr=v^2/r-g

I get something that looks like this:

dv/v=dr/r-dr/g ==> ln v = In r -r/g

Last edited: Sep 15, 2008
2. Sep 16, 2008

Redbelly98

Staff Emeritus
Hi,

What equations do you know of that deal with circular motion?

FYI, the "r" in m*M*G/r^2 is not necessarily a radius, it's the distance between two masses.

3. Sep 16, 2008

Benzoate

hi,

the equations that deal with circular motion are : F(centripetal)=m*v^2/r and the equation for a particle in general motion:
Well, since the airplane is obviously flying around a circle on earth, then doesn't the radius of the earth , plus the height the airplane is some distance above the surface of the earth become relevant in finding the total R=R(earth)+height?

4. Sep 16, 2008

tiny-tim

Hi Benzoate!

I'm really confused

the question gives you the acceleration (8g) … which you can assume is constant (they're telling you g = 10, which is about 2% out anyway )

you don't need G or M!

Try again (and don't forget you'll have to add or subtract up to 1g for the gravity)

5. Sep 16, 2008

Benzoate

could I just write:

a=v^2/r and since I know v and a I can easily calculate r correct?

6. Sep 16, 2008

Woohoo!