Solution:
Torque \(\tau = I\alpha ⇒ 3 = 1 \times \alpha ⇒ \alpha = 3\text{ rad/s}^2\). After 1 second, angular velocity is \(\omega = \alpha t = 3\text{ rad/s}\). The linear velocity is \(v = \omega R = 3 \times 1 = 3\text{ ms}^{-1}\).
A flywheel of moment of inertia \(1\text{ kg m}^2\) and radius 1 m starts rotating due to a constant torque 3 Nm. The velocity of a point on the rim after 1 s is (in \(\text{ms}^{-1}\))
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