Question

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

• A. $2.86kg{m}^2{s}^{-2}$
• B. $6.66kg{m}^2{s}^{-2}$
• C. $8.58kg{m}^2{s}^{-2}$
• D. $10.86kg{m}^2{s}^{-2}$

Answer 1

The correct option is B $6.66kg{m}^2{s}^{-2}$

Velocity of automobile, $v=54\times \frac{5}{18}=15m/s$
${\omega }_0=\frac{v}{r}=\frac{15}{0.45}=\frac{100}{3}rad/s$
Angular acceleration : $\alpha =\frac{0-{\omega }_0}{t}=-\frac{100}{45}$, since $t=15s$ as given.
Torque, $\tau =I\alpha =3\times \frac{100}{45}$$=6.66kg{m}^2{s}^{-2}$