Question

An automobile moves on a road with a speed of $54\, km\,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\,kg\,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

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

Answer 1

Answer: (C)

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