A constant torque of 100 N m turns a wheel of moment of inertia \(300 \text{kg} \text{m}^2\) about an axis passing through its centre. Starting from rest, its angular velocity after 3 s is
Solution:
Angular acceleration \(\alpha = \frac{\tau}{I} = \frac{100}{300} = \frac{1}{3} \text{rad/s}^2\). The angular velocity is \(\omega = \omega_0 + \alpha t = 0 + \left(\frac{1}{3}\right)(3) = 1 \text{rad/s}\).
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