A car accelerates uniformly from rest and reaches a speed of `27 m//s` in `9.00s`. If the diameter of a tire is `60 cm`, find (a) the number of revolutions the tire making during this motion, assuming that no slipping occurs. (b) What is the final rotational speed of a tire inrevolutions per second.

Answer

(a) `(405)/(2 pi)`
(b) `(45)/(pi) rev.sec`

Explanation

`u =0 v = 27 m//s t = 9 sec`
`v = u + at`
`27 = 0 + a xx 9`
`a = 3m//s^(2)` This is `ac c^(n)` of `COM`
As the wheel is in pure rolling
`because a = R alpha `
`3 = ((30)/(100)) alpha alpha = 10 rad//sec^(2)`
(a) `Delta theta = (1)/(2) alpha t^(2)`
`Delta theta = (1)/(2) xx 10 xx (9)^(2) = 405 rad`
No of revolution `= (Delta theta)/(2 pi) = (405)/(2 pi)`
(b) `omega = at`
`omega = 10 xx 9 = 90 rad//sec` or `omega = (45)/(pi) rev//sec`.