A man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is 75 % of that with outstretched arms, then his rotational kinetic energy will

Step-by-step solution

Let I₁ and ₁ represent the initial moment of inertia and angular velocity, with I₂ and ₂ the final values after the man folds his arms. The moment of inertia decreases to I₂ = 3 4 I₁ . Angular momentum conservation gives I₁ ₁ = I₂ ₂ . Substituting yields I₁ ₁ = 3 4 I₁ ₂ , so ₂ = 4 3 ₁ . The rotational kinetic energy increases from KE₁ = 1 2 I₁ ₁^2 to KE₂ = 1 2 I₂ ₂^2 = 1 2 ( 3 4 I₁ ) ( 4 3 ₁ )^2 = 1 2 I₁ ₁^2 4 3 = 4 3 KE₁ . The kinetic energy increases by KE = KE₂ - KE₁ = 1 3 KE₁ , representing a 1 3 100 % 33.3 % i