An automobile moves on a road with a speed of | vedclass

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(B) Given:Speed of the automobile,$v = 54 \,km h^{-1} = 54 	imes \frac{5}{18} \,m s^{-1} = 15 \,m s^{-1}$.Radius of the wheel,$R = 0.45 \,m$.Moment o

Vedclass · Accepted Answer

$6.66$

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(B) Given:Speed of the automobile,$v = 54 \,km h^{-1} = 54 	imes \frac{5}{18} \,m s^{-1} = 15 \,m s^{-1}$.Radius of the wheel,$R = 0.45 \,m$.Moment of inertia of the wheel,$I = 3 \,kg m^2$.Time taken to stop,$t = 15 \,s$.The initial angular speed of the wheel is $\omega_i = \frac{v}{R} = \frac{15 \,m s^{-1}}{0.45 \,m} = \frac{1500}{45} \,rad s^{-1} = \frac{100}{3} \,rad s^{-1}$.The final angular speed is $\omega_f = 0$ (as the vehicle comes to rest).The angular retardation of the wheel is $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - \frac{100}{3}}{15} = -\frac{100}{45} \,rad s^{-2}$.The magnitude of the average torque is $	au = I |\alpha| = 3 \,kg m^2 	imes \frac{100}{45} \,rad s^{-2} = \frac{300}{45} \,N m = \frac{20}{3} \,N m \approx 6.66 \,kg m^2 s^{-2}$.

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