# 7.4.3: Energy Equation in Rotating Coordinate System


The coordinate system rotating around fix axises creates a similar conservative potential as a linear system. There are two kinds of acceleration due to this rotation; one is the centrifugal and one the Coriolis force. To understand it better, consider a particle which moves with the our rotating system. The forces acting on particles are

$\label{ene:eq:rotatingF} \pmb{F} = \left( \overbrace{\omega^2 \,r \,\hat{r}}^{\text{ centrifugal}} + \overbrace {2\,\pmb{U} \times \boldsymbol{\omega} } ^{\text{Coriolis}} \right) \,dm$

The work or the potential then is

$\label{ene:eq:rotatingP} PE = \left( {\omega^2 \,r \,\hat{r}} + {2\,\pmb{U} \times \boldsymbol{\omega} }\right) \cdot d\boldsymbol{ll} \,dm$ The cylindrical coordinate are

$\label{ene:eq:cylinderical} d\boldsymbol{ll} = dr \hat{r} + r\,d\theta\, \hat{\theta} + dz\, \hat{k}$ where $$\hat{r}$$, $$\hat{\theta}$$, and $$\hat{k}$$ are units vector in the coordinates $$r$$, $$\theta$$ and $$z$$ respectively. The potential is then

$\label{ene:eq:rotatingPd} PE = \left( {\omega^2 \,r \,\hat{r}} + {2\,\pmb{U} \times \boldsymbol{\omega} }\right) \cdot \left( dr \hat{r} + r\,d\theta\, \hat{\theta} + dz\, \hat{k} \right)\,dm$ The first term results in $$\omega^2 \,r^2$$ (see for explanation in the appendix ?? for vector explanation). The cross product is zero of

$\nonumber \pmb{U} \times \boldsymbol{\omega} \times \pmb{U} = \pmb{U} \times \boldsymbol{\omega} \times \boldsymbol{\omega} = 0$ because the first multiplication is perpendicular to the last multiplication. The second part is

$\label{ene:eq:rotatingPdSecond} \left( {2\,\pmb{U} \times \boldsymbol{\omega} }\right) \cdot d\boldsymbol{ll}\,dm$ This multiplication does not vanish with the exception of the direction of $$\pmb{U}$$. However, the most important direction is the direction of the velocity. This multiplication creates lines (surfaces ) of constant values. From a physical point of view, the flux of this property is important only in the direction of the velocity. Hence, this term canceled and does not contribute to the potential. The net change of the potential energy due to the centrifugal motion is

$\label{ene:eq:NetPotentialR} PE_{centrifugal} = - \int_{1}^{2} \omega^2 \,r^2 \, dr \, dm = \dfrac{\omega^2 \left( {r_1}^2 - {r_2} ^2 \right) }{2} \,dm$ Inserting the potential energy due to the centrifugal forces into the energy equation yields

Energy Equation in Accelerated Coordinate

$\nonumber \dot{Q} - \dot{W} = \dfrac{d}{dt} \int_{cv} \left[ E_u + \dfrac{U^2}{2\dfrac{}{}} + a_x\,x + a_y\, y + (a_z + g) z - \dfrac{\omega^2 \,r^2}{2} \right] \rho\,dV \\ \label{ene:eq:ene:AccCVgeneral} + \int_{cv} \left( h + \dfrac{U^2}{2\dfrac{}{}} + a_x\,x + a_y\, y + (a_z + g)\,z - z\, \dfrac{\omega^2 \,r^2}{2} \right) U_{rn}\, \rho\,dA\\ \nonumber + \int_{cv} P\,U_{bn} \,dA$