Question

A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. Assuming the diameter of a tire is 58.0 cm, (a) find the number of revolutions the tire makes during this motion, assuming no slipping occurs, (b) What is the final angular speed of a tire in revolutions per second?

Answer 1

Given : v = 22 m/s , t = 9 s & d = 0.58 m

Solution :

(a) Since car starts from the rest , its initial velocity (u)= 0 ,

By equation of motion :

v = u + at

22= 0 + a(9)

i.e. a = 2.44 m/s²

Again , applying equation of motion :

S = ut +(1/2)at²

= 0 + (1/2)(2.44)(9)²

= 98.82 m

Circumference of the tire : πd = (3.14)(0.58)= 1.8212 m

Number of revolutions = 98.82/1.8212= 54.26 revolutions

(b)

Final velocity v = 22 m/s.

And rev per sec. = 22/1.8212 = 12.08 rev. per second

Angular speed( $\omega$ )= 2π×12.08 = 75.86 rad/s