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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA flywheel of mass \( M = 20 \text{ kg} \) and radius \( R = 0.2 \text{ m} \) (having moment of inertia \( I = \frac{1}{2}MR^2 \)) is mounted on a frictionless horizontal axle. A constant tangential pull of \( 25 \text{ N} \) is applied via a light cord wound around its rim. If the cord unwinds from rest over a length of \( 4 \text{ m} \), the final angular velocity \( \omega \) of the flywheel is:
Correct answer · Option 2
\( 22.4 \text{ rad s}^{-1} \)
Explanation
Work done \( W = F \times s = 25 \times 4 = 100 \text{ J} \). Moment of inertia \( I = \frac{1}{2}MR^2 = \frac{1}{2}(20)(0.2)^2 = 0.4 \text{ kg m}^2 \). By work-energy theorem, \( \frac{1}{2}I\omega^2 = 100 \implies \omega = \sqrt{\frac{200}{0.4}} = \sqrt{500} \approx 22.36 \text{ rad s}^{-1} \).
