Class 11 NEET Physics System of Particles and Rotational Motion

Here, Speed of the automobile,$v=54 \text{ kmh}^{-1}=54 \times \frac{5}{18} \text{ ms}^{-1}=15 \text{ ms}^{-1}$Radius of the wheel of the automobile, $R=0.45 \text{ m}$Moment of inertia of the wheel about its axis of rotation, $I=3 \text{ kg m}^2$Time in which the vehicle brought to rest, $t=15 \text{ s}$The initial angular speed of the wheel is$\omega_i=\frac{v}{R}=\frac{15 \text{ ms}^{-1}}{0.45 \text{ m}}=\frac{1500}{45} \text{ rads}^{-1}=\frac{100}{3} \text{ rads}^{-1}$and its final angular speed is $\omega_f=0$ (as the vehicle comes to rest) $\therefore$ The angular retardation of the wheel is$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{0-\frac{100}{3}}{15 \text{ s}}=-\frac{100}{45} \text{ rads}^{-1}$The magnitude of required torque is$\tau=I|\alpha|= (3 \text{ kg m}^2) \left(\frac{100}{45} \text{ rad s}^{-2}\right)$$=\frac{20}{3} \text{ kg m}^2 \text{ s}^{-2}=6.66 \text{ kg m}^2 \text{ s}^{-2}$