Acceleration in Circular Motion: \(\vec{a} = r\frac{d^2\theta}{dt^2}\hat{\theta} + r(\frac{d\theta}{dt})^2(-\hat{r})\)
For circular motion, show that \( \frac{d \hat{\theta}}{dt} = -\frac{d \theta}{dt}\hat{r}\)
We know that \( \hat{\theta } \) changes direction as we go around a circle, and so if a particle is undergoing circular motion, we expect \(\frac{d\hat{\theta }}{dt}\) to be non-zero. To figure out exactly what it is, let us write \(\hat{\theta }\) in terms of \(\hat{i}\) and \(\hat{j}\) for an arbitrary \(\theta\).
\( \begin{align} \hat{\theta} = -\sin(\theta) \hat{i} + \cos(\theta) \hat{j} \end{align} \)
Since \(\hat{i}\) and \(\hat{j}\) are constant anywhere in space, they are not time-dependent, so now the derivative becomes much more clear. Using the chain rule, we have:
\( \begin{align} \frac{d\hat{\theta}}{dt} = -\cos(\theta) \frac{d\theta}{dt} \hat{i} - \sin(\theta) \frac{d\theta}{dt} \hat{j} = - \frac{d\theta}{dt} \left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) \end{align} \)
We know that \( \left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) = \hat{r} \), so we can now simply write this derivative as:
\[ \begin{align} \frac{d\hat{\theta}}{dt} = - \frac{d\theta}{dt} \hat{r} \end{align} \]
