Question

A wheel whose moment of inertia is $0.03kg{m}^2$ is accelerated from rest to $20rad/s$ in $5s.$ When the external torque is removed, the wheel stops in $1$ min. Find $x$, when external torque is $x\times {10}^{-2}Nm$.

Answer 1

$\therefore {\tau }_{f}t=I\omega$
or ${\tau }_f=\frac{I\omega }{t}=\frac{0.03\times 20}{60}=0.01N.m$

During acceleration
$({\tau }_e-{\tau }_f)t=I\omega$
$\therefore {\tau }_e=\frac{I\omega }{t}+{\tau }_f=\frac{0.03\times 20}{5}+0.01$
$=13\times {10}^{-2}Nm$