72P - Fundamentals Of Physics

Question

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?

Step-by-Step Solution

Answer

Ratio of translational kinetic energy to rotational kinetic energy is 1.

Step 1: Given

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

Step 2: Determining the concept

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

Formulae are as follow:

 

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

Step 3: Determining the ratio of translational kinetic energy to rotational kinetic energy

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.