Simple Harmonic Motion of a car's tires

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SUMMARY

The discussion focuses on calculating the period of oscillation for a car tire with a hemispherical bump while traveling at a speed of 3.00 m/s. The relevant formula for the period of oscillation is T = 2π(I/(mgd))^0.5, where the moment of inertia (I) needs to be determined. The angular velocity of the tire is calculated as ω = v/R, resulting in an angular velocity of 10 rad/s for a tire radius of 0.30 m. The period of rotation is then derived using τ = 2π/ω, leading to a clear method for determining the oscillation period.

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  • Familiarity with angular velocity calculations
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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.
 
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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]
 
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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