## Centripetal Force Question

Hey guys -
Here is a problem I have been working on... I've tried several things, I think I'm pretty close...
Stone has a mass of 7e-3 kg and is wedged into the tread of a tire. Coefficient of static friction between each side of tread channel is 0.74. When the tire surface is rotating at 18 m/s the stone flies out. The magnitude of the normal force that each side of tread exerts is 1.8 N. Assume only static friction supplies centripetal force and determine radius of the tire.

So I have mass, coefficient of static friction, normal force, and speed.

I've tried...

Fc = ma_c = m(v^2/r)

F_c = f_s = u_sF_n
= 2(.74)2(1.8 N)

F_c = 5.3 N

r = F_c/(v2m)

5.3 N/(18 m/s)^2(7e-3) = 2.3 m

I've tried this and it came out wrong, of course thinking realistically it would be wrong anyways because usually tires (much less their radius) would be 2.3 m.

I could use F_c = 1.3 N (from not multiplying .74 by 2 and 1.8 N by 2)...

Any thoughts? Thanks!
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Recognitions: Homework Help A couple of comments. The friction force on one side is = µ*NR = 0.74*1.8 So, the friction from both sides is 2*µ*NR = 2*0.74*1.8 = 2.664 N. Also, Fc = mv²/r giving, r = mv²/Fc rather than Fc/mv²

 Similar discussions for: Centripetal Force Question Thread Forum Replies Classical Physics 18 Introductory Physics Homework 9 Introductory Physics Homework 6 General Physics 5 Introductory Physics Homework 8