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

Correct Answer: 3

The correct answer is 3.
$(KE{)}_{\text{Rotational}}=\frac{1}{2}I{\omega }^2=E$
$E=\frac{1}{2}\frac{m{\ell }^2}{12}{\omega }^2$
$E=\frac{1}{2}\frac{dA{\ell }^3}{12}{\omega }^2$
$E=\frac{dA(2{)}^3}{24}{\omega }^2$
$\sqrt{\frac{3E}{dA}}=\omega$
$\alpha =3$