Question

A flywheel of moment of inertia $3 \times 10^{2}\, kg - m ^{2}$ is rotating with uniform angular speed of $4.6\, rad / s$. 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, $I=3 \times 10^{2} kgm ^{2}$
Torque, $\tau=6.9 \times 10^{2} Nm$
Initial angular speed, $\omega_{0}=4.6\, rad\, s ^{-1}$
Final angular speed, $\omega_{0}=0\, rad\, s ^{-1}$
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, $\tau=I \alpha$
$6.9 \times 10^{2} =3 \times 10^{2} \times \frac{4.6}{t}$
$t =\frac{3 \times 10^{2} \times 4.6}{6.9 \times 10^{2}}=2\, s$