A 12.0 Nm torque is applied to a flywheel that rotates about a fixed axis and has a moment of inertia of 30.0 kgm2. If the flywheel is initially at rest, what is its angular velocity after it has turned through eight revolutions?
The flywheel turns through eight revolutions, which are 16π radians. The work done by the torque, which is constant and therefore can come outside the integral, is
\( \mathrm{W}_{\mathrm{AB}}=\tau\left(\theta_{\mathrm{B}}-\theta_{\mathrm{A}}\right) \)
We apply the work-energy theorem:
\( \mathrm{W}_{\mathrm{AB}}=\tau\left(\theta_{\mathrm{B}}-\theta_{\mathrm{A}}\right)=\frac{1}{2} I w_{B}^{2}-\frac{1}{2} I w_{A}^{2} \)
With τ = 12.0 N · m, θB − θA = 16.0 πrad, I = 30.0 kgm2 , and ωA = 0 , we have
Therefore, ωB = 6.3 rad/s. This is the angular velocity of the flywheel after eight revolutions.
