12.5: Newton’s Law of Motion in a Non-Inertial Frame

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The acceleration of the system in the rotating inertial frame can be derived by differentiating the general velocity relation for \(\mathbf{v}\), Equation \(12.4.4\), in the fixed frame basis which gives

\[\begin{align} \mathbf{a}_{fix} &= \left(\frac{d\mathbf{v}_{fix}}{dt}\right)_{fixed} \\[4pt] &= \left(\frac{d\mathbf{V}_{fix}}{dt}\right)_{fixed} + \left(\frac{d\mathbf{v}^{\prime\prime}_{rot}}{dt}\right)_{fixed} + \left(\frac{d\omega}{dt}\right)_{fixed} \times \mathbf{r}^{\prime}_{mov} + \omega \times \left(\frac{d\mathbf{r}^{\prime}_{mov}}{dt}\right)_{fixed} \label{12.22} \end{align}\]

Now we wish to use the general transformation to a rotating frame basis which requires inclusion of the time dependence of the unit vectors in the rotating frame, that is,

\[ \begin{align} \left(\frac{d\mathbf{v}^{\prime\prime}_{rot}}{dt}\right)_{fixed} &= \left(\frac{d\mathbf{v}^{\prime\prime}_{rot}}{dt}\right)_{rotating} + \omega \times \mathbf{v}^{\prime\prime}_{rot} \label{12.23} \\[4pt] \left(\frac{d\omega}{dt}\right)_{fixed} \times \mathbf{r}^{\prime}_{mov} &= \left(\frac{d\omega}{dt}\right)_{rot} \times \mathbf{r}^{\prime}_{mov} \label{12.24} \\[4pt] \omega \times \left(\frac{d\mathbf{r}^{\prime}_{mov}}{dt}\right)_{fixed} &= \omega \times \mathbf{v}^{\prime\prime}_{rot} + \omega \times (\omega \times \mathbf{r}^{\prime}_{mov}) \label{12.25} \end{align}\]

Using Equations \ref{12.23}, \ref{12.24}, \ref{12.25} gives

\[ \mathbf{a}_{fix} = \mathbf{A}_{fix} + \mathbf{a}^{\prime\prime}_{rot} + 2\omega \times \mathbf{v}^{\prime\prime}_{rot} + \omega \times (\omega \times \mathbf{r}^{\prime}_{mov}) + \dot{\omega} \times \mathbf{r}^{\prime}_{mov} \label{12.26}\]

where the acceleration in the rotating frame is \(\mathbf{a}^{\prime\prime}_{rot} = \left(\frac{d\mathbf{v}^{\prime\prime}_{rot}}{dt}\right)_{rot}\) while the velocity is \(\mathbf{v}^{\prime\prime}_{rot} = \left(\frac{\mathbf{r}^{\prime\prime}_{rot}}{dt}\right)_{rot}\) and \(\mathbf{A}_{fix}\) is with respect to the fixed frame.

Newton’s laws of motion are obeyed in the inertial frame, that is

\[ \begin{align} \mathbf{F}_{fix} &= m\mathbf{a}_{fix} \\[4pt] &= m(\mathbf{A}_{fix} + \mathbf{a}^{\prime\prime}_{rot} + 2\omega \times \mathbf{v}^{\prime\prime}_{rot} + \omega \times (\omega \times \mathbf{r}^{\prime}_{mov}) + \dot{\omega} \times \mathbf{r}^{\prime}_{mov}) \label{12.27} \end{align}\]

In the double-primed frame, which may be both rotating and accelerating in translation, one can ascribe an effective force \(\mathbf{F}^{eff}_{rot}\) that obeys an effective Newton’s law for the acceleration \(\mathbf{a}^{\prime\prime}_{rot}\) in the rotating frame

\[\begin{align}\mathbf{F}^{eff}_{rot} &= m\mathbf{a}^{\prime\prime}_{rot} \\[4pt] &= \mathbf{F}_{fix} - m(\mathbf{A}_{fix} + 2\omega \times \mathbf{v}^{\prime\prime}_{rot} + \omega \times (\omega \times \mathbf{r}^{\prime}_{mov}) + \dot{\omega} \times \mathbf{r}^{\prime}_{mov}) \label{12.28} \end{align}\]

Note that the effective force \(\mathbf{F}^{eff}_{rot}\) comprises the physical force \(\mathbf{F}_{fixed}\) minus four non-inertial forces that are introduced to correct for the fact that the rotating reference frame is a non-inertial frame.