Answer to Question #19420 in Mechanics | Relativity for Celesta
A boy on a bicycle, whose tire's radius is .5m, is traveling with a speed of 5m/s and begins braking. A bug clings tightly to the edge of the tire on the side. The tire coasts to a stop with angular acceleration of -.2rads/s^2. (a) How long did it take the bike, boy & bug to stop? (b) How many revolutions did the bug travel through before coming to a stop from th initial braking?
initial angular speed omega =v/R = (5 m/s)/(0.5 m) = 10 rad/s; if the braking time is T, then final speed = 0 = omega + acc*T = 10 rad/s - (0.2 rad/s^2)*T; T = (10 rad/s) / (0.2 rad/s^2) = 50 s.
The number of revolutions will be N = angle/(2*Pi) = (omega/2)*T = ( (10 rad/s)*(50 s)/2 ) / (2*Pi) = 39.81; Floored downwards, it's 39 revolutions.