Angular Impulse

Question

A constant torque of 5 N m is applied to a wheel of moment of inertia 2 kg m^2 that is initially at rest about its axis. What is the angular velocity attained by the wheel after the torque has acted for 4 seconds?

RankGuru · Accepted Answer

The angular impulse delivered by a torque equals the change in angular momentum, expressed as $	au \, 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 $\omega = \frac{	au \, t}{I} = \frac{(5)(4)}{2} = \frac{20}{2} = 10	ext{ rad/s}$. 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 $	au 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, $\frac{(	ext{N m})(	ext{s})}{	ext{kg m}^2} = \frac{	ext{kg m}^2 	ext{s}^{-2}\cdot	ext{s}}{	ext{kg m}^2} = 	ext{s}^{-1}$, 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 $\alpha = 	au/I = 2.5	ext{ rad/s}^2$ and then applying $\omega = \alpha t = 2.5 	imes 4 = 10	ext{ rad/s}$, so the impulse approach and the kinematic approach agree exactly, as they must when the torque is constant.

Angular Impulse

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