A thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

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:

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

Translational kinetic energy of pipe is given by following formula:

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

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,

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{rot}}=0.5\times \mathrm{I}\times {\mathrm{\omega }}^2 \\ {\mathrm{KE}}_{\mathrm{rot}}=0.5\times 0.5{\mathrm{mr}}^2\times {\mathrm{w}}^2 \\ {\mathrm{KE}}_{\mathrm{rot}}=0.5{\mathrm{mv}}^2 \end{gathered}$$

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.