Question

A uniform solid bar with mass m and length L rotates with angular velocity v about an axle at one end of the bar. What is the bar’s kinetic energy?

Short answer

The total kinetic energy of the bar is$\frac{1}{6}\left(m{\omega }^2{L}^2\right).$

Step-by-step solution

Content Introduction

The formula to find kinetic energy for the small element of bar is

$dK=\frac{1}{2}{v}^{2}dm$

here,$dK$is the kinetic energy of the small element of bar,$v$is linear velocity,$dm$is mass element of the same.

Content Explanation

Integrate the kinetic energy of each small element of the bar to find the total kinetic energy,

$$\begin{gathered} K={\int }_0^{L}dK \\ K={\int }_0^L\frac{1}{2}{v}^{2}dm \\ K={\int }_0^L\frac{1}{2}{\left(\omega x\right)}^{2}dm \end{gathered}$$

The formula to find the small mass is

$dm=\lambda dx$

Substitute

$$\begin{gathered} K={\int }_0^L\frac{1}{2}{\left(\omega x\right)}^{2}dm \\ K=\frac{1}{2}{\omega }^2\lambda {\int }_0^L{\left(x\right)}^{2}dx \\ K=\frac{1}{6}{\omega }^2\lambda L^3 \\ K=\frac{1}{6}m{\omega }^2{L}^2 \end{gathered}$$