- If the position vector of a particle in the cylindrical coordinates is $\mathbf{r}(t) = r\hat{\mathbf{e_r}}+z\hat{\mathbf{e_z}}$ derive the expression for the velocity using cylindrical polar coordinates.
I solved it this way $$\dot{\mathbf{r}}(t) = \dot{r}\hat{\mathbf{e_r}}+r\dot{\hat{\mathbf{e_r}}}+\dot{z}\hat{\mathbf{e_z}}+z\dot{\hat{\mathbf{e_z}}} = \dot{r}\hat{\mathbf{e_r}}+r\dot{\theta}\hat{\mathbf{e_\theta}}+\dot{z}\hat{\mathbf{e_z}}$$
As $\dot{\hat{\mathbf{e_z}}} =0$ and by finding the derivative of $\hat{\mathbf{e_r}}$ from the expression it in terms of $i$ and $j$.
- Consider the same situation as in the previous problem. Now assume, however, that the radial distance $r$ is constant ($\dot{r} = 0$). Derive the expression for the particle’s acceleration using the cylindrical coordinates.
$$\mathbf{\ddot{r}(t)} = \ddot{r}\hat{\mathbf{e_r}}+\dot{r}\dot{\hat{\mathbf{e_r}}}+\dot{r}\dot{\theta}\hat{\mathbf{e_\theta}}+r\ddot{\theta}\hat{\mathbf{e_\theta}}+r\dot{\theta}\dot{\hat{\mathbf{e_\theta}}}+\ddot{z}\hat{\mathbf{e_z}}+\dot{z}\dot{\hat{\mathbf{e_z}}} =(\ddot{r}-r\dot{\theta}^2)\hat{\mathbf{e_r}}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\mathbf{e_\theta}}+\ddot{z}\hat{\mathbf{e_z}} =-r\dot{\theta}^2\hat{\mathbf{e_r}}+r\ddot{\theta}\hat{\mathbf{e_\theta}}+\ddot{z}\hat{\mathbf{e_z}} $$
Do you think it's correct?
Best Answer
The expression you've got for the acceleration in cylindrical coordinates (for $r$ constant) is correct. Check here for similar derivations in other coordinate systems.