Question

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

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

Answer 1

The correct option is A $6.66{\text{kg m}}^2{\text{s}}^{-2}$

$v_i=15m/s$
$I=3kg\text{-}{m}^2$
$R=0.45m$
$t=15\text{s}$
$v_f=0$
$\tau =I\alpha$
${\omega }_f={\omega }_i-\alpha t$
$\alpha =\frac{{\omega }_i}{t}$

$v_i=R{w}_i$

$w_i=\frac{v_i}{R}=\frac{15}{0.45}=\frac{100}{3}$

$⟹\alpha =\frac{100}{3}\times \frac{1}{15}=\frac{100}{45}=6.66{\text{kg m}}^2{\text{s}}^{-2}$