Question

A flywheel of moment of inertia $3\times 10^{2}kg-m^2$ is rotating with uniform angular speed of $4.6rad/s$. If a torque of $6.9\times 10^{2}N-m$ retards the wheel, then the time in which the wheel comes to rest is

• A. 1.5 s
• B. 2 s
• C. 0.5 s
• D. 1 s

Answer 1

Answer: (B)

Here, moment of inertia, $I=3\times 10^{2}kg{m}^2$
Torque, $\tau =6.9\times 10^{2}Nm$
Initial angular speed, ${\omega }_0=4.6rad{s}^{-1}$
Final angular speed, ${\omega }_0=0rad{s}^{-1}$
As ${\omega }_0={\omega }_0+\alpha t$
$\therefore \alpha =\frac{\omega -{\omega }_0}{t}=\frac{0-4.6}{t}=-\frac{4.6}{t}rad{s}^{-2}$
Now, negative sign is for deceleration
torque, $\tau =I\alpha$
$6.9\times 10^2=3\times 10^2\times \frac{4.6}{t}$
$t=\frac{3\times 10^2\times 4.6}{6.9\times 10^2}=2s$