Question

A rod of mass $m$ and length $l$ is connected with a light rod of length $l$. The composite rod is made to rotate with angular velocity $\omega$ as shown in the figure. Then,

• A. total kinetic energy of the rod is $K_{total}=\frac{7}{6}m{\omega }^2{l}^2$
• B. translational kinetic energy, $K_{translational}=\frac{9}{8}m{\omega }^2{l}^2$
• C. rotational kinetic energy, $K_{rotational}=\frac{1}{24}m{\omega }^2{l}^2$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A total kinetic energy of the rod is $K_{total}=\frac{7}{6}m{\omega }^2{l}^2$
B translational kinetic energy, $K_{translational}=\frac{9}{8}m{\omega }^2{l}^2$
C rotational kinetic energy, $K_{rotational}=\frac{1}{24}m{\omega }^2{l}^2$

Translational kinetic energy of rod,
$K_{translational}=\frac{1}{2}m{v}_C^2$
Velocity of centre of mass $v_C=\omega \left(\frac{3}{2}l\right)$

Hence, $K_{transactional}=\frac{1}{2}m{\left[\frac{3\omega l}{2}\right]}^2=\frac{9}{8}m{\omega }^2{l}^2$

Rotational kinetic energy of rod $K_{rotational}=\frac{1}{2}{I}_c{\omega }^2=\frac{1}{2}\left[\frac{m{l}^2}{12}\right]{\omega }^2=\frac{1}{24}m{\omega }^2{l}^2$

Total Kinetic energy $K_{total}=K_{translational}+K_{rotational}$

$=\frac{9}{8}m{\omega }^2{l}^2+\frac{1}{24}m{\omega }^2{l}^2=\frac{7}{6}m{\omega }^2{l}^2$