Question

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 $\sqrt{\frac{\alpha Ad }{ Ad }}$ then value of $\alpha$ is _____

Answer 1

$ ( 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 $