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$\newcommand{\Prime}{{^\prime}}\newcommand{\pp}[2][]{\frac{\partial#1}{\partial #2}} \newcommand{\PP}[2][]{\frac{\partial^2#1}{\partial #2^2}} \newcommand{\dd}[2][]{\frac{d#1}{d #2}}\newcommand{\DD}[2][]{\frac{d^2#1}{d #2^2}}$
The equations of motion in a linearly accelerated are are derived and an expression for pseudo force is obtained.
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EOM in Non Inertial Frames
We will discuss the motion in a linearly accelerated frame. The examples are that of a lift, car or a train moving with a constant acceleration. I will always use symbol $K$ for an inertial frame. In most cases it will clear which frame is inertial. If it is not, we must identify an inertial frame. I assume that there is an observer $O$ who is performing an experiment and taking data. Then we have another frame $K\Prime$ accelerating with a constant acceleration. Let there be an observer $O\Prime$ in the frame $K\Prime$. Let $\vec{x}(t)$ and $\vec{x}\Prime(t)$ be the position vectors of a point at time $t$ as seen by the two observers $O$ and $O\Prime$ respectively. Let the positions of the axes in the two frames coincide at time $t=0$. Also let $\vec{x}_0(t)$ be the position vector of the origin of $K^\prime$ at time $t$ as seen by the observer $O$.
Then it is obvious that \begin{eqnarray} \vec{x}'(t)=\vec{x}_0(t) + \vec{x}\Prime(t) \end{eqnarray} Differentiating the above equation two times we get
\begin{eqnarray} \dd[\vec{x}(t)]{t}&=&\dd[\vec{x}_0(t)]{t} + \dd[\vec{x}^\prime(t)]{t}\\ \DD[\vec{x}(t)]{t}&=&\DD[\vec{x}_0(t)]{t} + \DD[\vec{x}^\prime(t)]{t} \end{eqnarray}
Therefore, the accelerations of a body, $\vec{a}$ and $\vec{a}\Prime(t)$, in the two frames are related by \begin{equation} \vec{a} = \vec{f}+\vec{a}\Prime(t) \end{equation} where $\vec{f}$ denotes the acceleration of the frame $K^\prime$ w.r.t. the frame $K$. Multiplying the above equation by the mass of the body
\begin{eqnarray} M \vec{a} &=& M \vec{f}+ M \vec{a}^\prime(t)\\ \text{or} \qquad \qquad M\vec{a}^\prime &=& M \vec{a}(t)- M \vec{f}\\ \end{eqnarray} Using Newton's Laws in the first frame $K$ and replacing $M\vec{a}(t)$ with force $\vec{F}$ we get \begin{equation}\label{EQ01} \text{or} \qquad \qquad M\vec{a}^\prime = \vec{F}- M \vec{f} \end{equation}
Pseudo forces
The second term in \eqref{EQ01}, $- M \vec{f} $, is called pseudo force. The EOM in the non inertial frame look similar but one has to include the pseudo force to the forces acting on the system. The pseudo forces to be included will depend on how the frame is accelerating. The pseudo force is equal to minus of mass times acceleration of the frame with respect to an inertial frame.
For rotating and other types of accelerated frames a similar approach is required to write the correct equations of motion. All pseudo forces have an important feature in common. They are proportional to the mass of the body and are independent of charge and other attributes. This is a property that is shared by the gravitational forces also.
