Question

An automobile traveling 87.0 km/h has tires of 68.0 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 19.0 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

The angular speed of the tyre about its axle is
$\omega$ = v / r
Where v is the velocity and r is the radius.
$\omega$ = 87.0 km/h / 68 x 10-2 m
$\omega$ = 87 x (103 / 3600) m/s /  (68/2) x 10-2 m
$\omega$ =  71.08 rad/s
b) The final angular velocity of the tyre become zero as it comes to a stop. It makes 19 complete turns before coming to halt.
$\omega$ f = 0
$\omega$ i = 71.08 rad/s
Using the relation,
$\omega$ f2 = $\omega$ i2 + 2 $\alpha$ $\theta$
Where $\theta$ is the angular position of the tyre in radians, $\alpha$ is the angular acceleration.
o = (71.08 rad/s)2 + 2 x $\alpha$ x 19 x 2 $\pi$
|$\alpha$| =   (71.08 rad/s)2 / 2 x19 x 2 $\pi$
|$\alpha$| =  21.17 rad/s2
c) The car moves a distance d as given by
d = $\theta$ r
Where r is the radius of the tyre.
d = 19 x 2$\pi$ x (68/2) x 10-2 m
d =  40.56 m