Question

An automobile moves on a rough road with a speed of $54{\text{kmh}}^{-1}$. 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 wheel is brought to rest in $15\text{s}$, then find magnitude of average torque transmitted by the brakes to wheel. Assume no slipping at all points of contact.

• A. $10.86{\text{kg m}}^2{\text{s}}^{-2}$
• B. $2.86{\text{kg m}}^2{\text{s}}^{-2}$
• C. $6.67{\text{kg m}}^2{\text{s}}^{-2}$
• D. $8.58{\text{kg m}}^2{\text{s}}^{-2}$

Answer 1

The correct option is C $6.67{\text{kg m}}^2{\text{s}}^{-2}$

Assuming the wheels are in state of pure rolling on the road and $v$ is the translational speed of $\text{COM}$ of the wheel.

$\therefore v=r\omega ...\left(i\right)$ ; condition of pure rolling
$r=0.45\text{m},v=54{\text{kmh}}^{-1}=15{\text{ms}}^{-1},I=3{\text{kg.m}}^2$


Since
$\tau =I\alpha =I\frac{d\omega }{dt}$
$\Rightarrow {\tau }_{avg}=I\times \frac{\Delta \omega }{\Delta t}...\left(ii\right)$
From Eq $\left(i\right)$ initially angular speed of wheel $\Rightarrow {\omega }_i=\frac{v}{r}=\frac{15}{0.45}$
Final angular speed of wheel after applying brakes ${\omega }_f=0$
$\Rightarrow$ From Eq $\left(ii\right)$:
${\tau }_{avg}=3\times \frac{(0-\frac{15}{0.45})}{15}$
$\therefore |{\tau }_{avg}|=\frac{100}{15}=6.66{\text{kg m}}^2{\text{s}}^{-2}$