Question

An automobile traveling 71.0 km/h has tires of 60.0 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 40.0 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? (Note: automobile moves without sliding)

Answer 1

Explanation:

We have,

Initial speed of an automobile, u = 71 km/h = 19.72 m/s

Diameter of the tie, d = 60 cm

Radius, r = 30 cm

(a) The angular speed of the tires about their axles is given by :

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{19.72}{0.3}\\\\\omega=65.73\ rad/s$

(b) Final angular velocity of the wheel is equal to 0 as its stops. The angular acceleration of the wheel is given by :

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

$\theta=40\ rev\\\\\theta=251.32\ rad$

$0-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_i^2}{2\theta}\\\\\alpha =\dfrac{(65.73)^2}{2\times 251.32}\\\\\alpha =-8.59\ rad/s^2$

(c) Let the car move a distance d during the braking. So,

$d=\theta r\\\\d=251.32\times 0.3\\\\d=75.39\ m$

Therefore, the above is the required explanation.