# Help with Centripetal Force question

1. Oct 11, 2008

### Sagekilla

1. The problem statement, all variables and given/known data
A 1000 kg sports car moving at 20 m/s crosses the rounded top of a hill (radius = 100m). Determine (a) the normal force on the car, (b) the normal force on the 70 kg driver, and (c) the car speed at which the normal force is zero.

2. Relevant equations
F = ma
Fn = mg
$$a = v^2/r$$

3. The attempt at a solution

I solved for acceleration:
a = (20 m/s)^2 / 100 m = 4 m/s^2

I then solved for the centripetal force:
F = 1000 kg * 4 m/s^2 = 4000 N

New work:

I know the acceleration due to centripetal force is +4 m/s^2, and it points inwards towards the hill. Since gravity is -10 m/s^2, the total acceleration is -6 m/s^2 into the hill.

I tried using this to solve for the normal force:
Fn = 1000 kg * -6 m/s^2 = -6000 N

(a) Fn = -6000 N

I then solved for the normal force, Fnd, on the 70 kg driver:
Fnd = 70 kg * -6 m/s^2 = -420 N

(b) Fnd = -420 N

To find the speed the car would have to move for the normal force to become zero, I would need the acceleration from gravity and acceleration from the centripetal force to cancel each other out:
Fn = 1000 kg * 0 m/s^2 = 0 N
a = -10 m/s^2 + 10 m/s^2

I found the force required to accelerate at 10 m/s^2:
F = ma = 1000 kg * 10 m/s^2 = 10000 N

Then I derived the velocity according to the acceleration formula:
v = sqrt(a * r)
v = sqrt(10 m/s^2 * 100m) = 31.62 m/s

(c) v = 31.62 m/s

Does this look correct?

Last edited: Oct 11, 2008
2. Oct 11, 2008

### Rake-MC

Looks fine to me. Most physics courses take gravity as 9.8 or 9.81 ms-2 but if you've been taking it as 10ms-2 then stick with that.