Question

A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s

Answer 1

(a) The car is undergoing an acceleration of

`a=(22.7(\rm m)/(\rm s)-0)/(9.02\,\mathrm s)\approx2.52(\rm m)/(\mathrm s^2)`

so that in 9.02 s, it will have covered a distance of

`x=\frac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m`

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

`(102\,\mathrm m)/(1.84\,\mathrm m)\approx55.7\,\mathrm{rev}`

(b) The wheels have average angular velocity

`\omega=\frac{\omega_f+\omega_i}2=(\theta_f-\theta_i)/(\Delta t)`

where
`\omega` is the average angular velocity,
`\omega_i` and
`\omega_f` are the initial and final angular velocities (rev/s),
`\theta_i` and
`\theta_f` are the initial and final angular displacements (rev), respectively, and
`\Delta t` is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

`\frac{\omega_f}2=(55.7\,\rm rev)/(9.02\,\rm s)\implies\omega_f\approx12.4(\rm rev)/(\rm s)`