Question

A man standing on a tum-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, his rotational kinetic energy will

विकल्प

• increase by 33.3%

• increase by 25%

• decrease by 25%

• decrease by 33.3%

Answer 1

increase by 33.3%

Explanation:

Let I₁ and ω₁ represent his moment of inertia and angular frequency when his arms are outstretched, and I₂ and ω₂ represent their values when his arms are folded. Then I₁ω₁ = I₂ω₂.

Now, I₂ = `3/4` I₁. Hence I₁ω₁ = `3/4` I₁ω₂ 

or ω₂ = `3/4` ω₁

Initial KE is K₁= `1/2` I₁ω₁² and final KE is

K₁= `1/2` I₁ω₂² = `1/2 xx (3I_1)/4 xx ((4ω_1)/3)^2`

= `4/3 (1/2 I_1ω_1^2) = 4/3`K₁

∴ Percentage increase in KE = `(K_2 - K_1)/K_1 xx 100`

= `((4/3 K_1 - K_1))/K_1 xx 100`

= `100/3` = 33.3%