An automobile moves on a road with a speed of $54 \mathrm{~km} \mathrm{~h}^{-1}$. The radius of its

Question

An automobile moves on a road with a speed of $54 \mathrm{~km} \mathrm{~h}^{-1}$. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is $3 \mathrm{~kg} \mathrm{~m}^{2}$. If the vehicle is brought to rest in 15 s , the magnitude of average torque transmitted by its brakes to the wheel is\

Professor PC Thomas Classes · Answer

(c) : Here, speed of the automobile,
$ v=54 \mathrm{~km} \mathrm{~h}^{-1}=54 	imes \frac{5}{18} \mathrm{~m} \mathrm{~s}^{-1}=15 \mathrm{~m} \mathrm{~s}^{-1} $
 Radius of the wheel of the automobile, $R=0.45 \mathrm{~m}$\ Moment of inertia of the wheel about its axis of rotation,\ $I=3 \mathrm{~kg} \mathrm{~m}^{2}$\ Time in which the vehicle brought to rest, $t=15 \mathrm{~s}$\ The initial angular speed of the wheel is\ $\omega_{i}=\frac{v}{R}=\frac{15 \mathrm{~m} \mathrm{~s}^{-1}}{0.45 \mathrm{~m}}=\frac{1500}{45} \mathrm{rad} \mathrm{s}^{-1}=\frac{100}{3} \mathrm{rad} \mathrm{s}^{-1}$\ and its final angular speed is $\omega_{f}=0$\ (as the vehicle comes to rest)\ $	herefore \quad$ The angular retardation of the wheel is 
$ \alpha=\frac{\omega_{f}-\omega_{i}}{t}=\frac{0-\frac{100}{3}}{15 \mathrm{~s}}=-\frac{100}{45} \mathrm{rad} \mathrm{~s}^{-2} 
$ 
The magnitude of required torque is 
$ \begin{aligned} & 	au=I|\alpha|=\left(3 \mathrm{~kg} \mathrm{~m}^{2}ight)\left(\frac{100}{45} \mathrm{rad} \mathrm{~s}^{-2}ight) \ & =\frac{20}{3} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}=6.66 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} \end{aligned} $