Question #e264b

Jan 21, 2017

Velocity of the automobile $v = 80 \text{km/h"=(80xx10^3m)/(60xx60s)=200/9"m/s}$

Radius of Tyre $r = 80 c m = 0.8 m$

Initial angular velocity of wheels

${\omega}_{0} = \frac{v}{r} = \frac{200}{0.8 \times 9} \text{rad/s"=250/9"rad/s}$

On application of brakes the car stops in 30 complete turns. So the angular displacement of wheels $\theta = 30 \times 2 \pi = 60 \pi$ radian.
Final angular velocity of wheel $\omega = 0$

If angular acceleration be $\alpha$ then we can write from equation of rotational motion

${\omega}^{2} = {\omega}_{0}^{2} + 2 \alpha \times \theta$

Inserting values we get

${0}^{2} = {\left(\frac{250}{9}\right)}^{2} + 2 \times \alpha \times 60 \pi$

$\implies \alpha = - \frac{250 \times 250}{9 \times 9 \times 120 \pi} \approx - 2 \text{rad/} {s}^{2}$