[NOTES/ME-08002]-Acceleration in a rotating frame

1. Statement of the problem

Imagine that the two observers \(O{'},O{''}\) are observing a body moving under influence of some force \(\mathbf F\). The Newton's laws of motion are applicable in an inertial frame and cannot be used for writing the equation of motion in a rotating frame. However we are not interested in solving the EOM in the inertial frame, the solution is required for the observer in the rotating frame. For this purpose we need the equation of motion in the rotating frame of reference.

A word about vector notation: We use boldface letter to denote a vector, for example \(\mathbf A\). The components of a vector with respect to different sets of axes will be denoted by \(\vec{A}{'}=(A_1{'}, A_2{'}, A_3{'})\), \(\vec{A}{''}=(A_1{''}, A_2{''}, A_3{''})\).

2.Relating Accelerations in inertial and rotating frames

In order to set up the equation correctly we need to relate the acceleration in the two frames. We shall do this in several steps here. Let us take a point \(A\) fixed in frame \(K{''}\). Since the frame \(K{''}\) is rotating w.r.t. \(K{'}\), the observer \(O{'}\) will see the vector changing with time, in fact rotating w.r.t. the frame \(K{'}\). I want to compute its time derivative as seen by the observer \(O{'}\). The components of \(\bf A\) will be related as \begin{equation} \vec{A}{'}(t)=\vec{A}{''} + \sin \theta (\hat{n}\times \vec{A}{''}) + (1-\cos\theta) \hat{n}\times(\hat{n}\times \vec{A}{''}) \end{equation} where \(\theta\) depends on time and \(\theta=\theta(t)=\omega t\). We will need the following identity \begin{equation} \hat{n}\times\vec{A}{'}(t)=\sin \theta \hat{n}\times(\hat{n}\times \vec{A}{''})+ \cos\theta(\hat{n}\times \vec{A}{''}) \end{equation} The time derivative, as computed by the observer \(O{'}\), will be given by \begin{eqnarray} \dd{t} \vec{A}{'}(t)\nonumber &=&\cos \theta(t) \omega (\hat{n}\times \vec{A}{''}) + \sin\theta(t))\omega \hat{n}\times(\hat{n}\times \vec{A}{''})\\ &=& \omega \hat{n}\times \vec{A}{'}(t) = \vec{\omega}\times \vec{A}{'}(t).\label{EQ03} \end{eqnarray} This result could also have been derived using the result on change in a vector \(\Delta \vec{A}{'} = \Delta \theta (\hat{n}\times \vec{A}{'})\) under an infinitesimal \eqRef{EQ03} applies to any vector which is fixed in frame \(K{''}\). In particular it applies to the unit vectors \(\hat{u}_k,k=1,2,3\) along the coordinate axes in \(K{''}\). So get \begin{equation} \dd{t} \hat{u}_k = \vec{\omega}\times \hat{u}_k \end{equation} Let us now take up the general case of a vector \(\mathbf A\) which is varying with time in the frame \(K{''}\). So for example, the vector \(\mathbf A\) could be position, velocity or acceleration of a moving particle under influence of some forces. For a such a case we would get \begin{eqnarray} \dd [{\mathbf A}]{t} &=& \dd[A_k{''}]{t} \hat{u}_k + A_k{''} \dd[\hat{u}_k]{t}.\\ &=& \dd[A_k{''}]{t} \hat{u}_k + A_k{''}\big(\vec{\omega}\times \hat{u}_k\big)\label{EQ07} \\ &=& \dd[A_k{''}]{t} \hat{u}_k + \vec{\omega}\times {\mathbf A}. \end{eqnarray} Differentiating \eqRef{EQ07} once again we would get \begin{eqnarray}\label{EQ08} \DD [{\mathbf A} ]{t} &=& \dd{t}\left(\dd[A_k{''}]{t} \hat{u}_k + A_k{''} \dd[\hat{u}_k]{t}\right).\\ &=& \left(\DD[A_k{''}]{t}\right) \hat{u}_k +\left(\dd[A_k{''}]{t} \right)\left(\dd[\hat{u}_k]{t}\right) + \left(\dd[A_k{''}]{t}\right) \big(\vec{\omega}\times \hat{u}_k\big) + A_k{''} \dd{t} \Big(\vec{\omega}\times \hat{u}_k\Big)\label{EQ09}\\ &=& \left(\DD[A_k{''}]{t}\right) \hat{u}_k + 2 \vec{\omega}\times \hat{u}_k \dd[A_k{''}]{t} + A_k{''} \big(\vec{\omega}\times(\vec{\omega}\times\hat{u}_k)\big). \end{eqnarray} Now we identify \(K{'}\) with an inertial frame, \(\mathbf A\) with position vector \(\mathbf x\) in rotating frame \(K{''}\) and write the above result as \begin{eqnarray}\label{EQ08A} \DD [{\mathbf x} ]{t} &=& \left(\DD[x_k]{t}\right) \hat{u}_k + 2 \vec{\omega}\times \hat{u}_k \Big(\dd[x_k]{t} \Big)+ x_k \big(\vec{\omega}\times(\vec{\omega}\times\hat{u}_k)\big). \end{eqnarray} The left hand side is just the acceleration of the body as seen by the observer in the inertial frame. Hence we would get \begin{eqnarray}\label{EQ10} \frac{\mathbf F}{m} &=& \left(\DD[x_k]{t}\right) \hat{u}_k + 2 \vec{\omega}\times \hat{u}_k \Big(\dd[x_k]{t} \Big)+ x_k \big(\vec{\omega}\times(\vec{\omega}\times\hat{u}_k)\big). \end{eqnarray} Remembering that the unit vectors \(\hat{u}_k\) do not change with time in the rotating frame, the velocity and acceleration, \({\mathbf v}_r, {\mathbf a}_r \) in the rotating frame are give by \begin{eqnarray} {\mathbf v}_r\equiv\dd[{\mathbf x}]{t}\Big|_\text{rf} &=& \dd[(x_k \vec{u_k})]{t}\Big|_\text{rf} = \vec{u_k}\dd[x_k ]{t}\Big|_\text{rf} \\ [2mm] {\mathbf a}_r\equiv\DD[{\mathbf x}]{t}\Big|_\text{rf} &=& \DD[(x_k \vec{u_k})]{t}\Big|_\text{rf} = \vec{u_k}\DD[x_k ]{t}\Big|_\text{rf}. \end{eqnarray} where the quantities \({\mathbf v}\) and \({\mathbf a}\) denote the velocity and acceleration of the body as seen in the rotating frame. Thus we get \begin{equation} m {\mathbf a}_r = {\mathbf F} -2 \vec{\omega}\times {\mathbf v}_r - \vec{\omega}\times( \vec{\omega}\times {\mathbf x}). \end{equation} We may write this equation of motion in the rotating frame as a differential equation \begin{equation}\label{EQ18} m \DD[\vec{x}]{t} = \vec {F} -2 \vec{\omega}\times \dd[\vec{x}]{t} - \vec{\omega}\times( \vec{\omega}\times \vec{x}). \end{equation} Here it is understood that the quantities in \eqRef{EQ18}, viz. \( \vec{x}, \vec{v} \) and the time derivatives, refer to the rotating frame. This lecture is based on Kleppner and Kolenkow, \S 8.5 \cite{Kleppn}