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Centripetal Force Question

  1. Oct 12, 2005 #1
    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!
  2. jcsd
  3. Oct 13, 2005 #2


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    Homework Helper

    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


    r = mv²/Fc rather than Fc/mv²
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