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6.System of Particles and Rotational Motion
medium
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 $\omega$. If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\frac{\alpha E}{ Ad }}$ then the value of $\alpha$ is $...........$
A
$2$
B
$1$
C
$4$
D
$3$
(JEE MAIN-2023)
Solution
$( KE )_{\text {Rotational }}=\frac{1}{2} I \omega^2= E$
$E =\frac{1}{2} \frac{ m \ell^2}{12} \omega^2$
$E =\frac{1}{2} \frac{ dA \ell^3}{12} \omega^2$
$E =\frac{ dA (2)^3}{24} \omega^2$
$\sqrt{\frac{3 E }{ dA }}=\omega$
$\alpha=3 \text { Ans. }$
Standard 11
Physics
