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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA rigid system of total mass \( M = 4 \text{ kg} \) moves in a plane. Its centre of mass moves with speed \( V = 3 \text{ m s}^{-1} \). The system simultaneously rotates about an axis passing through its centre of mass with angular speed \( \omega = 5 \text{ rad s}^{-1} \). If the moment of inertia about this axis is \( I = 0.8 \text{ kg m}^2 \), the total kinetic energy of the system according to the relation \( K = K' + \frac{1}{2}MV^2 \) is:
View Answer & Solution
Correct answer · Option C
\( 28 \text{ J} \)
Explanation
\( K_{\text{trans}} = \frac{1}{2}MV^2 = \frac{1}{2}(4)(3)^2 = 18 \text{ J} \). Internal kinetic energy \( K' = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.8)(5)^2 = 10 \text{ J} \). Total \( K = 10 + 18 = 28 \text{ J} \).
