Angular Impulse

The angular impulse delivered by a torque equals the change in angular momentum, expressed as $\tau t=\Delta L=I\Delta \omega$, the rotational counterpart of the linear impulse-momentum theorem. Since the wheel starts from rest, $\Delta \omega =\omega$, so $\omega =\frac{\tau t}{I}=\frac{(5)(4)}{2}=\frac{20}{2}=10\text{ rad/s}$. The value 40 rad/s forgets to divide by the moment of inertia. The value 2.5 rad/s omits the time factor and divides torque by inertia alone. The value 20 rad/s reports the angular impulse $\tau t$, which has the units of angular momentum, rather than the angular velocity itself. This applies the NCERT relation between torque, time, and the change in angular momentum directly. As a unit check, $\frac{(\text{N m})(\text{s})}{{\text{kg m}}^2}=\frac{{\text{kg m}}^2{\text{s}}^{-2}\cdot \text{s}}{{\text{kg m}}^2}={\text{s}}^{-1}$, the dimension of angular velocity, confirming both the method and the final answer of 10 rad/s. The same result follows from first computing the angular acceleration $\alpha =\tau /I=2.5{\text{ rad/s}}^2$ and then applying $\omega =\alpha t=2.5\times 4=10\text{ rad/s}$, so the impulse approach and the kinematic approach agree exactly, as they must when the torque is constant.