A rod of length $L$ revolves in a horizontal plane about the axis passing through its centre and perpendicular to its length. The angular velocity of the $\operatorname{rod}$ is $\omega$. If $A$ is the area of cross-section of the rod and $\rho$ is its density, then the rotational kinetic energy of the rod is

• A

  $\frac{1}{3} A L^3 \rho \omega^2$
• B

  $\frac{1}{2} A L^3 \rho \omega^2$
• Correct answer
  C

  $\frac{1}{24} A L^3 \rho \omega^2$
• D

  $\frac{1}{18} A L^3 \rho \omega^2$

Solution

Rotational kinetic energy of rod is given as
$K_{\mathrm{rot}}=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{M L^2}{12}\right) \omega^2$ $K_{\mathrm{rot}}=\frac{1}{24} M L^2 \omega^2$ $\ldots(\mathrm{i})$ But $\quad M=$ volume of rod $\times$ density $=A L \times \rho$ $M=A L \rho$ $\ldots(\mathrm{ii})$ From Eqs. (i) and (ii), we have $K_{\mathrm{rot}}=\frac{1}{24} \times A L \rho \times L^2 \omega^2=\frac{1}{24} A L^3 \rho \omega^2$

Hence, the correct answer is option: 'C'.