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

#### intdx

1. The problem statement, all variables and given/known data
From http://library.thinkquest.org/10796/index.html
4. A 1300 kg car takes 35 s to travel around a circular road that has a radius of 40 m.
...
b. What is the acceleration of the car?
c. How big should the force of friction be to maintain this acceleration?
d. How big should the coefficient of the force of friction be?
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.

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#### cepheid

Staff Emeritus
Gold Member
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.

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