Finding the Force Acting on a Particle Moving Along an Ellipse

The position of a particle of mass \(m\) that is moving along an ellipse with a constant angular speed \(\omega\) is given by the vector function \begin{equation*} \mathbf{r}(t)=A\cos ( \omega t) \mathbf{i}+B\sin ( \omega t) \mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Find the force \(\mathbf{F}\) acting on the particle at any time \(t.\) Graph \(\mathbf{r}=\mathbf{r}( t)\) and \(\mathbf{F}=\mathbf{F}( t) .\)

Solution To find the force \(\mathbf{F}\), we use Newton's Second Law of Motion, \(\mathbf{F}=m\mathbf{a}\). We begin by finding the acceleration \(\mathbf{a}\) of the particle. \begin{eqnarray*} \mathbf{r}(t)&\;=\;&A\cos ( \omega t) \mathbf{i}+B\sin ( \omega t) \mathbf{j} \\[3pt] \mathbf{v}(t)&\;=\;&\mathbf{r^{\prime} }(t)=-A\omega \sin ( \omega t) \mathbf{i}+B\omega \cos ( \omega t) \mathbf{j} \\[3pt] \mathbf{a}(t)&\;=\;&\mathbf{r^{\prime \prime} }(t)=-A\omega ^{2}\cos ( \omega t) \mathbf{i}-B\omega ^{2}\sin ( \omega t) \mathbf{j} =-\omega ^{2}\mathbf{r}(t) \end{eqnarray*}

Then, by Newton's Law, \begin{equation*} \mathbf{F}(t)=m\mathbf{a}(t)=-m\hbox{ }\omega ^{2}\mathbf{r}(t) \end{equation*}

The direction of the force vector \(\mathbf{F}\) is opposite to that of the vector \(\mathbf{r}\) at any time \(t\), as shown in Figure 27.