What is the outer radius of a car tire that has tangential speed of 36m/s and a period of rotation of 5.60*10e-2?

If the period of rotation, T,  of the car tire is `5.6*10^(-2) ` seconds, then the frequency of the rotation, f, can be found as

`f = 1/T = 17.9 s^(-1)`

The angular speed of the tire, `omega` , relates to the frequency as

`w=2pif = 112.2` rev/s (revolutions per second.)

If the tangential speed of the tire is 36 m/s, it means that a point on the outside of the tire (a point on the rim of the tire) moves with the linear velocity of 36 m/s. This point has the radius of rotation equal to the outer radius of the tire, r.

The relationship between the linear velocity and the angular velocity is

`v = omega*r`

Then, the outer radius of the tire is

`r = v/omega = 36/112.2 =0.32 m` .

The outer radius of the car tire is 0.32 meters, which is 32 centimeters, or approximately 1 ft.