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 is √((α E)/(Ad)) then value of α is ______________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Mass of rod m = A d L = 2Ad (with L = 2 m). I = (mL²)/(12) = (2Ad × 4)/(12) = 2Ad/3 E = 1/2Iω² = (Ad ω²)/(3) ⇒ ω = √(3E/Ad) So α = 3.

A thin uniform rod of length 2 m, cross sectional area ' A ' and… | JEE Main 2023 (30 Jan, Shift 1)

Q107JEE Main 2023 · 30 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A thin uniform rod of length $2\text{ m}$ , cross sectional area ' $A$ ' and density ' $d$ ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$ . If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\frac{\alpha E}{Ad}}$ then value of $\alpha$ is ______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Mass of rod $m = A d L = 2Ad$ (with $L = 2\text{ m}$ ).
$I = \frac{mL^2}{12} = \frac{2Ad \times 4}{12} = \frac{2Ad}{3}$
$E = \frac{1}{2}I\omega^2 = \frac{Ad\,\omega^2}{3} \Rightarrow \omega = \sqrt{\frac{3E}{Ad}}$
So $\alpha = 3$ .

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