91Ó°ÊÓ

Axis of rotation is passing through the center about its length.

Use formulas for kinetic energy for translation and rotational motion then take their ratio.

Formulae are as follow:

$$\begin{gathered} {\mathbf{KE}}_{\mathbf{trans}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{×}\mathbf{m}\mathbf{×}{\mathbf{v}}^{\mathbf{2}} \\ {\mathbf{KE}}_{\mathbf{rot}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{×}\mathbf{I}\mathbf{×}{\mathbf{Ó\neg }}^{\mathbf{2}} \end{gathered}$$

Translational kinetic energy of pipe is given by following formula:

${\mathrm{KE}}_{\mathrm{trans}}=0.5×\mathrm{m}×{\mathrm{v}}^2$

To find rotational kinetic energy about center parallel to length, inertia about center of pipe is needed which is given as follow:

$\mathrm{I}={\mathrm{mr}}^2$

And, rotational kinetic energy is given as,

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{rot}}=0.5×\mathrm{I}×{\mathrm{Ó\neg }}^2 \\ {\mathrm{KE}}_{\mathrm{rot}}=0.5×0.5{\mathrm{mr}}^2×{\mathrm{w}}^2 \\ {\mathrm{KE}}_{\mathrm{rot}}=0.5{\mathrm{mv}}^2 \end{gathered}$$

So, ratio is given as,

$\frac{{\mathrm{KE}}_{\mathrm{trans}}}{{\mathrm{KE}}_{\mathrm{rot}}}=\frac{0.5{\mathrm{mv}}^2}{0.5{\mathrm{mv}}^2}$

So,

$\frac{{\mathrm{KE}}_{\mathrm{trans}}}{{\mathrm{KE}}_{\mathrm{rot}}}=1$

Hence,ratio of translational kinetic energy to rotational kinetic energy is 1.

Therefore, using the formula for the rotational and translational kinetic energy, the ratio of these quantities can be found.