Classical Mechanics

10.1 Circular Motion – Acceleration

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Acceleration in Circular Motion: $\overrightarrow{a}=r\frac{d^2\theta }{d{t}^2}\hat{\theta }+r(\frac{d\theta }{dt}{)}^2(-\hat{r})$

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 $\theta$.

$\begin{array}{r}\hat{\theta }=-\sin (\theta )\hat{i}+\cos (\theta )\hat{j}\end{array}$

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:

$\begin{array}{r}\frac{d\hat{\theta }}{dt}=-\cos (\theta )\frac{d\theta }{dt}\hat{i}-\sin (\theta )\frac{d\theta }{dt}\hat{j}=-\frac{d\theta }{dt}(\cos (\theta )\hat{i}+\sin (\theta )\hat{j})\end{array}$

We know that $(\cos (\theta )\hat{i}+\sin (\theta )\hat{j})=\hat{r}$, so we can now simply write this derivative as:

$$\begin{array}{r}\frac{d\hat{\theta }}{dt}=-\frac{d\theta }{dt}\hat{r}\end{array}$$

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