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A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Its kinetic energy is

1. 1/2 m·r²·ω²
2. m·r²·ω²
3. m·r²·ω²
4. 1/2 m·r·ω²

Correct answer: 1/2 m·r²·ω²

Solution

The rotational kinetic energy of a body is given by (1/2)Iω², where I is the moment of inertia. For a ring rotating about its central axis, I = m·r². Substituting this, the kinetic energy becomes (1/2)m·r²·ω².

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