Question

A motorcycle is moving with speed $20\text{m/s}$ on the road. The M.O.I of wheel about its axis of rotation is $2{\text{kg m}}^2$ and radius of its wheel is $0.5\text{m}$. Motorcycle comes to rest in $10\text{sec}$ after applying the breaks. Find the magnitude of the average torque transmitted by its brakes to the wheel?

• A. $8{\text{kg m}}^2{\text{/s}}^2$
• B. $16{\text{kg m}}^2{\text{/s}}^2$
• C. $4{\text{kg m}}^2{\text{/s}}^2$
• D. $2{\text{kg m}}^2{\text{/s}}^2$

Answer 1

The correct option is A $8{\text{kg m}}^2{\text{/s}}^2$

Given velocity of motorcycle, $v=20\text{m/s}$
So, initial angular velocity $w_o=\frac{v}{r}=\frac{20}{0.5}$
$w_o=40\text{rad/sec}$
Now, $w=w_0+\alpha t$
Final angular velocity $w=0$ as motorcycle comes in rest in $10\text{sec}$.
$0=40+\alpha \left(10\right)$ (given $t=10\text{sec})$
$\alpha =-4\text{rad/sec}$
The magnitude of average torque,
$\left|\tau \right|=I\left|\alpha \right|=2\times 4=8{\text{kg m}}^2{\text{/s}}^2$
$\left|\tau \right|=8{\text{kg m}}^2{\text{/s}}^2$