Question

Dimensional analysis of Kinetic energy of a Rotating body

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Kinetic}\mathrm{energy}\mathrm{of}\mathrm{rotating}\mathrm{body},{\mathrm{K}}_{\mathrm{r}}=\frac{1}{2}{\mathrm{I\omega }}^2 \\ \mathrm{where},\mathrm{I}=\mathrm{moment}\mathrm{of}\mathrm{inertia} \\ \mathrm{\omega }=\mathrm{angular}\mathrm{velocity} \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{I}=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^0\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{\omega }=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}{\mathrm{K}}_{\mathrm{r}}=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^0\right]\times {\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]}^2 \\ \mathrm{Here},\mathrm{we}\mathrm{are}\mathrm{not}\mathrm{considering}\left(\frac{1}{2}\right)\mathrm{because}\mathrm{it}\mathrm{is}\mathrm{constant}. \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}{\mathrm{K}}_{\mathrm{r}}=\left[{\mathrm{M}}^{1+0}{\mathrm{L}}^{2+0}{\mathrm{T}}^{0-2}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}{\mathrm{K}}_{\mathrm{r}}=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}\right] \end{gathered}$$