[NOTES/CM-08004] Equation of Motion in Non Inertial Frames

Equation of motion in a rotating frame

Let us consider motion of a particle as seen by an observer in a rotating frame $K_r$ Let $K_i$ be an inertial frame. In the frame $K_i$, Newton's Laws will be valid and the equations of motion can be written down. Let $\vec{F}_e$ denote the external force acting on the body and let \(\vec X\) denote the position of the body in the inertial frame. Then the equations of motion in frame \(K_i\) are of the standard form
\begin{equation} m \ddot {\vec X} =\vec F_e \end{equation}
For a conservative system \(\vec F_e=-\nabla V\). These equations can now be transformed and written in terms of the position of the body w.r.t the non-inertial frame $K$. Instead of following this route, we start with the Lagrangian
\begin{eqnarray} L&=&\frac{1}{2}m \dot {\vec X}^2 - V(X)\Label{EQ110A}\\ &=&\frac{1}{2}m{\U{\dot X}^T}\U{\dot X}-V(\vec{\U{X}}),\Label{EQ110B} \end{eqnarray}
Here \(\U{X}\) stands for the position in the matrix notation
\begin{equation}
\U{X} = \begin{pmatrix} X_1\\X_2\\X_3\end{pmatrix}.
\end{equation}
We will write the Lagrangian, \eqref{EQ110B}, in terms of the position of the body as seen by an observer in the rotating frame $K_r$. We assume $K_r$ is rotating about an axis $\hat{n}$ with uniform angular velocity $\omega$ w.r.t the frame \(K_i\). Then angle of rotation at time $t$ is \(\theta=\omega t, (\dot{\theta}=\omega), \) and the coordinates, \(\U{x}\), of the body in the frame \(K_r\) are given by
\begin{equation} \U{x} =\U{R}_{\hat{n}}(\theta) \U{X} \end{equation}
\where $R_{\hat{n}}(\theta)$ is the rotation matrix. \begin{equation} \U{R}_{\hat{n}}(\theta)=\exp(-\theta\hat{n}\cdot\vec{I}). \end{equation} We use
\begin{equation} \U{X}=R_{\hat{n}}^{T} \U{x}=\exp(\theta\hat{n}\cdot\vec{I})\U{x}. \end{equation}
to compute time variation of \(\U{X}(t)\):
\begin{eqnarray} \frac{d}{dt} \U{X}&=&\Big\{\frac{d}{dt}\exp(\theta\hat{n}\cdot\vec{I})\Big\} \U{x}+\exp(\theta\hat{n}\cdot\vec{I})\frac{d\U{x}}{dt} \\ &=&e^{\theta \hat{n}\cdot\vec{I}}\Big[\frac{d\U{x}}{dt}+\big(e^{-\theta\vec{I}\hat{n}}\frac{d}{dt } e^ { \theta\hat{n}\cdot\vec{I}}\big)\U{x}\Big] \\ &=&R^{T}\big(\dot{\U{x}}+(\vec{\omega}\cdot{\vec{I}})\U{x}\big) \end{eqnarray}
Therefore, we have a useful result

\begin{equation}\boxed{\dot{\vec{X}}=R^T(\dot{\vec{x}}+\vec{\omega}\times\vec{x})}\end{equation}

This result can be derived in several ways. It holds for the rates of change of every time dependent vector in the two frames. We can now transform the Lagrangian \eqref{EQ110B} and write it in terms of the coordinates and velocities in the non inertial frame.
\begin{eqnarray} L&=&\frac{1}{2}m{\U{\dot X}^T}\U{\dot X}-V(\vec{\U{X}})\\ &=&\frac{1}{2}m(\dot{\vec{x}}\times(\vec{\omega}\times\vec{x}))^2-V(\vec{x}) \end{eqnarray}
The Euler Lagrange equations for \(x_k\) are
\begin{equation}\Label{EQ12} \dd{t}\Big(\pp[L]{\dot x_k}\Big) -\pp[L]{x_k}=0 \end{equation}
\For purposes of computing derivatives we proceed as follows. Let \(\hat e_k, k=1,2,3\) denote the unit vectors along the coordinate axes of the frame \(K_r\). We write
\begin{eqnarray} \vec x &=& \sum_k x_k \hat e_k, \qquad \pp[\vec x]{x_k} = \hat e_k,\\ \dot{\vec x} &=& \sum_k \dot x_k \hat e_k, \qquad \pp[\dot{\vec x}]{\dot x_k} = \hat e_k. \end{eqnarray}
We now compute the derivatives required in Euler Lagrange equation \eqref{EQ12}. %
\begin{eqnarray} \frac{\partial L}{\partial \dot{x_k}} &=&\frac{m}{2} \pp{\dot {x_k}} (\dot{\vec{x}}+\vec{\omega}\times\vec{x})^2\\ &=&m(\dot{\vec{x}}+(\omega\times\vec{x})).\pp{\dot x_k}(\dot{\vec{x}}+(\omega\times\vec{x})) \\ &=&m(\dot{\vec{x}}+(\omega\times\vec{x})).\hat e_k\\ &=&m(\dot{\vec{x}}_k+(\omega\times\vec{x})_k) \end{eqnarray}

Differentiating w.r.t time we get
\begin{equation} \dd{t}\Big(\frac{\partial L}{\partial \dot{x_k}} \Big)=m(\ddot{\vec{x}}_k+(\omega\times\dot{\vec x})_k) \end{equation}

Similarly,
\begin{eqnarray} \frac{\partial L}{\partial {x_k}} &=&\frac{m}{2} \pp{{x_k}}(\dot{\vec{x}}+\vec{\omega}\times\vec{x})^2\\ &=&m(\dot{\vec{x}}+\vec{\omega}\times\vec{x}) \pp{{x_k}}(\dot{\vec{x}}+\vec{\omega}\times\vec{x})\\ &=&m[\dot{\vec{x}}\cdot(\omega\times\hat{e}_k)+(\omega\times\vec{x}).(\vec \omega \times \hat e_k)]\\ &=&m[(\dot{\vec{x}}\times \omega)\cdot\hat{e}_k) +((\omega\times\vec{x})\times\vec \omega) \cdot \hat e_k] \\ &=&m[-(\omega \times\dot{\vec{x}})\cdot\hat{e}_k -(\vec \omega\times(\omega\times\vec{x}) )\cdot \hat e_k] \end{eqnarray}
Therefore, from \EqRef{EQ12}, we get
\begin{eqnarray} m\ddot{\vec{x}}_k+2m(\omega\times{\dot{\vec{x}}})_k+m(\vec{\omega}\times(\vec{\omega} \times\vec{x}))_k+\frac{\partial V}{\partial{x}_k}&=&0. \end{eqnarray}
We can now write the EOM as a vector equation
\begin{eqnarray} m\ddot{\vec{x}}=\vec{F_{e}}-2m\vec{\omega}\times{\dot{\vec{x}}}-m\vec{\omega} \times(\vec{\omega}\times\vec{x}) \end{eqnarray}
where \(\vec {F}_e\) is the external force. Thus, as seen from the rotating frame the particle moves as if it is under additional pseudo forces, known as the centrifugal force and the Coriolis force.

• $-2m\vec{\omega}\times{\dot{\vec{x}}}$ is called Coriolis force
• $-m\vec{\omega}\times(\vec{\omega}\times\vec{x})$ is known as centrifugal force.

It may be noted that the Coriolis force vanishes if the body is at rest in the rotating frame.