Q.A constant torque of $12\ \text{N}\,\text{m}$ is applied to a wheel of moment of inertia $3\ \text{kg}\,\text{m}^2$, initially at rest. Find

(a) Angular velocity after $5\ \text{s}$: first find the angular acceleration from $\tau=I\alpha$:

Then, starting from rest ($\omega_0=0$):

(b) Rotational kinetic energy gained:

Verification using work done by the torque: the angle turned through in this time is $\theta = \omega_0t + \tfrac12\alpha t^2 = 0 + \tfrac12(4)(5)^2 = 0.5(4)(25) = 50\ \text{rad}$, so the work done is

$W = \tau\theta = 12(50) = 600\ \text{J}$ …

Find $\alpha=\tau/I$, then $\omega=\alpha t$ from rest; separately compute $KE_{rot}=\tfrac12I\omega^2$ and cross-check it ag …

• Forgetting to square $\omega$ in the kinetic energy formula.
• Using the given time $t=5\ \text{s}$ directly as $\theta$ in $W=\tau\theta$, instead of first co …