Question

An automobile moves on a road with a speed of $54km{h}^{-1}$.The radius of its wheels 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 transmitted by its brakes to the wheel is

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

Answer 1

Answer: (C)

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