Question

An automobile traveling 55.0 km/h has tires of 77.0 cm diameter.
(a) What is the angular speed of the tires about their axles?
(b) If the car is brought to a stop uniformly in 38.0 complete turns of the tires, what is the magnitude of the angular acceleration of the wheels?
(c) How far does the car move during the braking?

Answer 1

Explanation:

Given that, Speed of the automobile, v = 55 km/h = 15.27 m/s

Diameter of the tire, d = 77 cm

Radius, r = 0.385 m

(a) Let $\omega$ is the angular speed of the tires about their axles.

The relation between the linear speed and the angular speed is given by :

$v=r\times \omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{15.27\ m/s}{0.385\ m}$

$\omega=39.66\ rad/s$

(b) Number of revolution,

$\theta=38\ rev=238.76\ radian$

Final angular speed of the car, $\omega_f=0$

Initial angular speed, $\omega_i=39.66\ rad/s$

Let $\alpha$ is the angular acceleration of the car. Using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\alpha =\dfrac{-\omega_i^2}{2\theta}$

$\alpha =\dfrac{-(39.66)^2}{2\times 238.76}$

$\theta=-3.29\ rad/s^2$

(c) Let d is the distance covered by the car during the braking. It is given by :

$d=\theta\times r$

$d=238.76\times 0.385$

d = 91.92 meters