The moment of inertia of a thin ring about an axis through its center and perpendicular to its plane is I_ring = M R². The moment of inertia of a thin disc about an axis through its center and perpendicular to its plane is I_disc = (1/2) M R². Since I_ring > I_disc (MR² is greater than (1/2)MR²), and ω is the same for both, the rotational kinetic energy of the ring is higher than that of the disc (KE_ring = (1/2)I_ringω² > KE_disc = (1/2)I_discω²).
A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their center of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true ?
Asked in UPSC NDA 2 2019 · Topic: Mechanics (Rotational Dynamics and Moment of Inertia)
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