Question

A rod of mass $m$ spins with an angular speed $\omega =\sqrt{g/l}$, Find its
a. Kinetic energy of rotation.
b. kinetic energy of translation
c. Total kinetic energy.

Answer 1

(a) kinetic energy of rotation $=\frac{1}{2}{I}_{CM}{w}^2$

$K{E}_{rot}=\frac{1}{2}\left(\frac{1}{3}m{l}^2\right){\left(\sqrt{\frac{g}{g}}\right)}^2=\frac{1}{2}\times \frac{1}{3}m{l}^2\times \frac{g}{l}=\frac{mgl}{6}$

(b) kinetic energy of translation $=\frac{1}{2}m{V}_{CM}^2$

$K{E}_{tran}=\frac{1}{2}\times m\times {\left(\frac{l}{2}w\right)}^2=\frac{1}{2}\times m\times \frac{l^2}{4}\times {\left(\sqrt{\frac{g}{l}}\right)}^2=\frac{mgl}{8}$

$K{E}_{tran}=\frac{1}{8}mgl$

(c) Total kinetic energy $=K{E}_{rot}+K{E}_{tran}$

$K{E}_{total}=\frac{1}{6}mgl+\frac{1}{8}mgl$

$K{E}_{total}=(\frac{1}{6}+\frac{1}{8})mgl$

$K{E}_{total}=\frac{7}{24}mgl$