Question

A fly wheel of moment of inertia $3\times {10}^{2}kg{m}^2$ is rotating with uniform angular speed of 4.6 $rad{s}^{-1}.$ If a torque of $6.9\times {10}^2$ Nm 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
• E. 2.5 s

Answer 1

The correct option is B 2 s

Here, $I=3\times {10}^{2}kg{m}^2$
$\tau =6.9\times {10}^{2}Nm$
${\omega }_0=4.6rad{s}^{-1}$
$\omega =0rad{s}^{-1}$
Now, $\tau =I\alpha$
$6.9\times {10}^2=3\times {10}^2\times \frac{4.6}{t}$
$t=\frac{3\times {10}^2\times 4.6}{6.9\times {10}^2}=2s.$