# Homework Help: Centripetal accel - airplane flying in a horizontal circle

1. Nov 6, 2007

### lettertwelve

centripetal accel -- airplane flying in a horizontal circle...

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

An airplane is flying in a horizontal circle at a speed of 116 m/s. The 77.0 kg pilot does not want his centripetal acceleration to exceed 6.00 times free-fall acceleration.

(a) What is the minimum radius of the circular path? (in meters)

(b) At this radius, what is the net force that maintains circular motion exerted on the pilot by the seat belts, the friction between him and the seat, and so forth?

2. Relevant equations

i'm more concentrated on part A.
this is what i have so far:

F=ma=m(v^2/R)=mG

v=116m/s
G=6

so then v^2/R=6, 116^2/R=6

so then
R=2242.7

it says my answer is incorrect.
i don't see my mistake though.

and for part B, would it be 0?

PART A: SOLVED. but i still need help with part B.
Fnet=ma, but i dont think that one works in this case...

Last edited: Nov 6, 2007
2. Nov 6, 2007

### Astronuc

Staff Emeritus
but wait, the problem states "6.00 times free-fall acceleration", which means an acceleration of 6g or 6 * 9.81 m/s2, not simply 6.

3. Nov 6, 2007

### lettertwelve

ahh yes! WOW i must be blind. thanks!

now for part B...

4. Nov 6, 2007

### GTrax

Does it not seem a tad unlikely that a pilot doing a turn of 2.247 km radius (thats more than 1.2 miles!) would be pulling a body-bending 6g ?
Is the 9.81m/s/s value of g included in the given information you have to solve this?
(OK - I see he got there, but some answers do make for great mental pictures of extreme stuff )

Last edited: Nov 6, 2007
5. Nov 6, 2007

### lettertwelve

for g, instead of 9.8 we can just use 10.

but anyways i do see what you mean :)

6. Nov 6, 2007

anyone?? :'[

7. Nov 6, 2007

### rl.bhat

Pilot cannot fly in the horizontal circle unless he tilts the plane toward the center. The angle of tilt can be found by tan(theta) = v^2/Rg. Then you can find the rest of the values.

Last edited: Nov 6, 2007
8. Nov 7, 2007

### lettertwelve

he didnt give us any angles

9. Nov 7, 2007

### rl.bhat

In part a you have found the velocity by using 6g = v^2/R.
The angle of tilt can be found by tan(theta) = v^2/Rg.
Hence v^2/R = g*tan(theta). Substitute in the first equation, we get 6g = g*tan(theta)
or tan(theta) =6 or theta = tan^-1(6) = 80.5 degree.
When the pilot is flying with this tilt, the normal component of his weight(R) = mgcos(theta) = 77*10*cos(80.5) N. Centripetal force (F)= 6g = 60N. mu = F/R