Question

An automobile traveling 89.0 km/h has tires of 62.0 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 27.0 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

Answer:

a) 79.7rad/s

b) -18.7rad/s^2

c) 53m

Explanation:

We will use the MKS system of unit, so:

$v=89.0km/h=89.0\frac{km}{h}*\frac{1000m.h}{3600km.s}=24.7m/s\\\\d=62.0cm=62.0cm*\frac{0.01m}{1cm}=0.62m$

now, The angular speed is given by:

$\omega=\dfrac{v}{\frac{d}{2}}\\\\\\\omega=\frac{24.7m/s}{0.31m}=79.7rad/s$

in order to obtain the angular acceleration we have to apply the following formula:

$(\omega_f)^2=(\omega_o)^2+2\alpha*\theta\\\\\alpha=-\frac{(\omega_o)^2}{2*rev*2\pi}\\\\\alpha=-\frac{(79.7m/s)^2}{2*27*2\pi}=-18.7rad/s^2$

The linear displacement is given by:

$d_l=\theta*r\\d_l=rev*2\pi*\frac{d}{2}\\\\d_l=27*2\pi*0.31m=53m$