Simple Harmonic Motion of a car's tires

AI Thread Summary
To determine the period of oscillation for the bump on the car's tire, start by calculating the angular velocity using the formula ω = v/R, where v is the car's speed (3.0 m/s) and R is the tire's radius (0.30 m), resulting in an angular velocity of 10 rad/s. The period of rotation can then be found using τ = 2π/ω, which gives the time it takes for one complete rotation of the tire. The moment of inertia is not directly needed for this calculation, as the focus is on the relationship between linear speed and angular motion. Understanding these concepts allows for the determination of the bump's oscillation period effectively. The calculations lead to a clearer understanding of simple harmonic motion in this context.
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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 \omega = v/R = (3.0m/s)/0.30m = 10 rad/s The period of rotation then is then \tau = \frac{2\pi}{\omega}
 
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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