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A uniform rod PQ of length l is rotating in a horizontal plane about the axis YY'. If the rod has cross-sectional area A, density ρ, and angular speed ω, then its rotational kinetic energy is

1. (1/3) A l³ ρ ω²
2. (1/2) A l³ ρ ω²
3. (1/24) A l³ ρ ω²
4. (1/18) A l³ ρ ω²

Correct answer: (1/24) A l³ ρ ω²

Solution

For rotation about the perpendicular axis YY' through the rod's centre, I = (1/12) M l^2 with M = A l rho. KE = (1/2) I w^2 = (1/2)(1/12)(A l rho) l^2 w^2 = (1/24) A l^3 rho w^2.

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