Proof that the acceleration vector always points radially inward for a body in circular motion
A few days ago, me and a friend of mine challenged each other to see if we could mathematically prove that the acceleration vector always points radially inward for a body in circular motion.
So here's what I came up with:
Proof
We know the position vector for a body making an angle \(\theta_0\) with the X-axis is given by:
\begin{align*} \vec{r}=r\,(\cos{(\theta_0)}\,\hat{i}+\sin{(\theta_0)}\,\hat{j}) \\ \end{align*} But for a body revolving around the origin, the angle changes at the rate \(\omega\). That is to say:
\begin{align*} \omega=\frac{d\theta}{dt} \\ \end{align*} So in general, the angle at any given time \(t\) would be (\(\theta_0+\omega t\)). Which implies the position vector would be:
\begin{align*} \vec{r}=r\,(\cos{(\theta_0+\omega t)}\,\hat{i}+\sin{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Now since we have the position as a function of time, we can take the its derivative w.r.t. time to get the velocity function.
\begin{align*} \vec{v}=\frac{d\vec{r}}{dt}=r\omega\,(-\sin{(\theta_0+\omega t)}\,\hat{i}+\cos{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Taking the derivative again w.r.t. time, gives us the acceleration function.
\begin{align*} \vec{a}=\frac{d\vec{v}}{dt}=-r\omega^2\,(\cos{(\theta_0+\omega t)}\,\hat{i}+\sin{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Now it is trivial to see that the coefficient of \(\omega^2\) in the above equation is \(\vec{r}\), the position function we started off with. So after substitution, we get:
\begin{align*} \vec{a}=-\vec{r}\omega^2 \\ \end{align*}
Conclusion: The equation above implies that the acceleration vector is proportional to negative of the position vector. And since we know that the position vector always points radially outward from the origin, we can conclusively say that the acceleration vector would always points radially inward.
So here's what I came up with:
Proof
\begin{align*} \vec{r}=r\,(\cos{(\theta_0)}\,\hat{i}+\sin{(\theta_0)}\,\hat{j}) \\ \end{align*} But for a body revolving around the origin, the angle changes at the rate \(\omega\). That is to say:
\begin{align*} \omega=\frac{d\theta}{dt} \\ \end{align*} So in general, the angle at any given time \(t\) would be (\(\theta_0+\omega t\)). Which implies the position vector would be:
\begin{align*} \vec{r}=r\,(\cos{(\theta_0+\omega t)}\,\hat{i}+\sin{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Now since we have the position as a function of time, we can take the its derivative w.r.t. time to get the velocity function.
\begin{align*} \vec{v}=\frac{d\vec{r}}{dt}=r\omega\,(-\sin{(\theta_0+\omega t)}\,\hat{i}+\cos{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Taking the derivative again w.r.t. time, gives us the acceleration function.
\begin{align*} \vec{a}=\frac{d\vec{v}}{dt}=-r\omega^2\,(\cos{(\theta_0+\omega t)}\,\hat{i}+\sin{(\theta_0+\omega t)}\,\hat{j}) \\ \end{align*} Now it is trivial to see that the coefficient of \(\omega^2\) in the above equation is \(\vec{r}\), the position function we started off with. So after substitution, we get:
\begin{align*} \vec{a}=-\vec{r}\omega^2 \\ \end{align*}
Conclusion: The equation above implies that the acceleration vector is proportional to negative of the position vector. And since we know that the position vector always points radially outward from the origin, we can conclusively say that the acceleration vector would always points radially inward.
Q.E.D.
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