# Centripetal Force of park ride

1. Oct 8, 2005

### rockmorg

I have a question on what it is I might be doing wrong here... I've tried this answer in the grader and it is wrong...

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s and an 83 kg person feels a 495 N force pressing against his back. Radius of the chamber?

Here is what I have done..

v = 3.2 m/s
m = 83 kg
F = 495 N
r = ?

Fc = mv2/r
r = Fc(1/mv2)

r = 495 N (1/(83kg * 3.2 m/s2))
r = .582 m

Even just looking at it it looks wrong, heh...

Any help would be great!

Thanks,
-
Morgan

2. Oct 8, 2005

### Pyrrhus

Notice this:

$$F = m \frac{v^2}{r}$$

$$\frac{r}{F} F = \frac{r}{F} m \frac{v^2}{r}$$

$$r = m \frac{v^2}{F}$$

3. Oct 8, 2005

### hotvette

I was just going to say that it looks like you didn't square the velocity.

4. Oct 8, 2005

### rockmorg

Ahhh thanks much - so I guess it was just a matter of arranging my equation? The mass needs to multiply out to those other variables for it to work...