JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Rotational Mechanics

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A thin uniform rod of length 2 m, cross sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity

ω

. If value of

ω

in terms of its rotational kinetic energy

E

\frac{α E}{A d}

then the value of

α

____

Answer: 3

Solution

The rod's rotational energy is

E = \frac{1}{2} I ω^{2}

with

I = \frac{m ℓ ^{2}}{12}

and mass

m = d A ℓ

. For

ℓ = 2 m

I = \frac{d A ( 2 ) ^{3}}{12} = \frac{2 d A}{3}

, so

E = \frac{1}{2} \cdot \frac{2 d A}{3} ω^{2} = \frac{d A}{3} ω^{2}

. Solving for the angular velocity,

ω = \frac{3 E}{A d}

, which on comparison with

\frac{α E}{A d}

gives

α = 3

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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