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Simple Harmonic Motion of a car's tires

  1. Jan 10, 2005 #1


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    While riding behind a car traveling at 3.00 m/s, you notice that one of the car's tires has a small hemispherical bump on it's rim. If the radii of the car's tires are .300 m, what is the bump's period of oscillation?

    Can anyone tell me where to start with this? I'm actually not even sure what formula to use. I think I have to use T=2pi(I/(mgd))^.5
    I don't know how to calculate the moment of inertia in that, though.
  2. jcsd
  3. Jan 10, 2005 #2
    Since you are traveling 3m/s, the car in front of you is also traveling that fast. Since the rotating tires are giving rise to the car's velocity, the tire's angular velocity is given by [tex]\omega = v/R = (3.0m/s)/0.30m = 10 rad/s[/tex] The period of rotation then is then [tex]\tau = \frac{2\pi}{\omega}[/tex]
  4. Jan 10, 2005 #3
    You know the linear speed and the radii of the tyre. So, first the angular velocity of the tyre. Then work out the period.
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