# Centripetal Force: finding accel., force of frict., and coeff. of the force of frict.

1. Jul 1, 2012

### intdx

1. The problem statement, all variables and given/known data
From http://library.thinkquest.org/10796/index.html
m=1300kg
t=35s
r=40m
2. Relevant equations
$$V=\frac{2\pi r}{t}$$
$$a_c=\frac{V^2}{r}$$
$$F_c=ma_c$$
$$F_f=\mu mg$$
$$F_f=F_c$$
$$\mu=\frac{a_c}{g}=\frac{F_c}{mg}$$
3. The attempt at a solution
Plugging into the formulae is pretty elementary, so I'll spare you all that. I used the first two equations for part b, then the third for part c. Anyway, I get 1.289m/s2 for part b and 1675.819N for part c, but the site's answers are 1.47m/s2 and 2210N, respectively. It appears that, to have arrived at their answer for part b, they used the incorrect equation a_c=V^2/t. I'm not sure how they arrived at the answer for part c.

Using either part of the last equation above (with my answers for part b or part c) gives me μ=0.132. The site's answer for part d is 0.15, but--using the same equation that I have listed above--this is consistent with its answer for part b, but not with its answer for part c.

Would someone please explain what's going on here?

Thanks!!

P.S. You know, I feel as if, from typing this, I've pretty much assured myself that the site is wrong. (I have another fairly recent post in which I consider the same thing, and the couple of replies it's gotten seem to support the idea.) I hope that doesn't sound arrogant, but it bothers me that the site was written by high school students.

2. Jul 2, 2012

### cepheid

Staff Emeritus
Re: Centripetal Force: finding accel., force of frict., and coeff. of the force of fr

I agree with the numbers you got. I think that the site is wrong.