Physics - Rotation Question with Solution | TestHub

Mass of the rod, $M=\text{Volume}\times \text{density}=A\times L\times d$

Given $L=2$, $M=2Ad$

Moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is $I=\frac{M{L}^2}{12}$

Substituting the values of $M$ and $L$, we get $I=\frac{2Ad\times 2^2}{12}=\frac{2Ad}{3}$

Rotational kinetic energy is given by $E=\frac{1}{2}I{\omega }^2$

Substituting the value of $I$, $E=\frac{1}{2}\left(\frac{2Ad}{3}\right){\omega }^2=\frac{Ad}{3}{\omega }^2$

Rearranging for $\omega$, we get $\omega ={\left(\frac{3E}{Ad}\right)}^{1/2}$

Comparing this with the given expression $\omega ={\left(\frac{\alpha E}{Ad}\right)}^{1/2}$, we find $\alpha =3$.

Answer: $3$