- #1
Rct33
- 16
- 0
Homework Statement
I am asked to find the radial and transverse velocity and acceleration for a particle with polar coordinates [itex]r=e^t[/itex] and [itex]\theta=t[/itex]
The Attempt at a Solution
Therefore let the position of the particle be [itex]\underline{r}=\underline{\hat{r}}e^t[/itex]
$$\therefore \dot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$
But [itex]\underline{\hat{r}}=\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}}[/itex]
$$\therefore\dot{\underline{r}}=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}})e^t+e^t\underline{\hat{r}}$$
$$=(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t +e^t\underline{\hat{r}}$$
[itex]\therefore \underline{\hat{r}}=e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}[/itex]
This is the particles velocity, the acceleration is given by:
$$\ddot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}}e^t+\underline{\hat{r}}e^t)$$
$$=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{\theta}}+\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$
$$=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t+e^t \underline{\hat{\theta}}+e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$$
$$=-e^t\underline{\hat{r}}+e^t\underline{\hat{\theta}}+e^t \underline{\hat{\theta}}+e^t\underline{\hat{r}}$$
$$=2e^t\underline{\hat{\theta}}$$
This is an acceleration in the transverse direction. But if [itex]\theta=t[/itex] and [itex]r=e^t[/itex] then why do I get an acceleration in the transverse direction, and not the radial direction? Is the transverse velocity not constant (I also got [itex]e^t\underline{\hat{\theta}}[/itex] for transverse velocity, why is this?) Help is very much appreciated