Velocity and Acceleration in a Coordinate System
Spherical Coordinates: Velocity and Acceleration
$$ \begin{align*} \mathbf{v} &=\dot{r} \hat {\mathbf{r}} +r \dot{\theta} \hat{ \boldsymbol{\theta}}+ r \dot{\phi} \sin{\theta} \hat{ \boldsymbol{\phi}} \\ \mathbf{a} &= (\ddot{r}-r\dot\theta^2-r\dot\phi^2\sin^2\theta)\hat{\mathbf{r}}+(r\ddot\theta+2\dot{r}\dot\theta-r\dot\phi^2\sin\theta\cos\theta)\hat{\boldsymbol{\theta}} \\ &\quad+(r\ddot\phi\sin\theta+2\dot{r}\dot\phi\sin\theta+2r\dot\theta\dot\phi\cos\theta)\hat{\boldsymbol{\phi}} \end{align*} $$
Derivation
Unit vectors in spherical coordinates are as follows.
$$ \begin{align*} \hat{\mathbf{r}} &= \cos \phi \sin \theta \hat{\mathbf{x}} + \sin \phi \sin \theta \hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\theta}} &= \cos\phi \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta \hat{\mathbf{y}} - \sin\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\phi}} &= -\sin\phi \hat{\mathbf{x}} + \cos\phi \hat{\mathbf{y}} \end{align*} $$
Let’s now find the velocity and acceleration in spherical coordinates. Velocity is derived by differentiating position over time, and acceleration is derived by differentiating velocity over time. Note that $\dot{r}$ is read as “al dot.” In physics, a dot over a letter signifies differentiation with respect to time.
$$ \dot{r}=\frac{dr}{dt} $$
Before laying out the velocity and acceleration, let’s precompute the differentiation of unit vectors. $\dot{\hat{ \mathbf{r}}}$, $\dot {\hat{ \boldsymbol{\theta}}}$, $\dot {\hat{ \boldsymbol{\phi}}}$ each yield the following.
$$ \begin{align*} \dot{\hat{ \mathbf{r}}} &= \frac{d}{dt}(\cos\phi\sin\theta\hat{\mathbf{x}} +\sin\phi\sin\theta\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} ) \\ &= \frac{d\cos\phi}{dt}\sin\theta\hat{\mathbf{x}} + \cos\phi\frac{d \sin\theta}{dt}\hat{\mathbf{x}}+\frac{d \sin\phi}{dt}\sin\theta\hat{\mathbf{y}} + \sin\phi\frac{d\sin\theta}{dt}\hat{\mathbf{y}}+\frac{d \cos\theta}{dt}\hat{\mathbf{z}} \\ &= \frac{d\cos\phi}{d\phi}\frac{d\phi}{dt}\sin\theta\hat{\mathbf{x}} + \cos\phi\frac{d \sin\theta}{d\theta}\frac{d\theta}{dt}\hat{\mathbf{x}} \\ &\quad +\frac{d \sin\phi}{d\phi}\frac{d\phi}{dt}\sin\theta\hat{\mathbf{y}} + \sin\phi\frac{d\sin\theta}{d\theta}\frac{d\theta}{dt}\hat{\mathbf{y}}+ \frac{d \cos\theta}{d\theta}\frac{d\theta}{dt}\hat{\mathbf{z}} \\ &= -\dot\phi\sin\phi \sin\theta\hat{\mathbf{x}} + \dot\theta\cos\phi\cos\theta\hat{\mathbf{x}}+\dot\phi\cos\phi \sin\theta\hat{\mathbf{y}} + \dot\theta\sin\phi\cos\theta\hat{\mathbf{y}}-\dot\theta\sin\theta\hat{\mathbf{z}} \\ &= \dot{\theta}(\cos\phi\cos\theta\hat{\mathbf{x}} + \sin\phi\cos\theta\hat{\mathbf{y}} -\sin\theta \hat{\mathbf{z}}) + \dot{\phi}\sin\theta (-\sin\phi\hat{\mathbf{x}}+\cos\phi\hat{\mathbf{y}}) \\ &= \dot{\theta} \hat{ \boldsymbol{\theta} } + \dot{\phi} \sin\theta \hat{ \boldsymbol{\phi}} \end{align*} $$
$$ \begin{align*} \dot {\hat{ \boldsymbol{\theta}}} &= \frac{d}{dt}(\cos\phi\cos\theta\hat{\mathbf{x}} + \sin\phi\cos\theta\hat{\mathbf{y}} -\sin\theta \hat{\mathbf{z}}) \\ &= -\dot{\phi}\sin\phi\cos\theta\hat{\mathbf{x}}-\dot\theta\cos\phi\sin\theta\hat{\mathbf{x}}+\dot{\phi} \cos\phi \cos\theta \hat{\mathbf{y}}-\dot{\theta} \sin\phi \sin\theta \hat{\mathbf{y}}-\dot\theta\cos\theta\hat{\mathbf{z}} \\ &= -\dot\theta (\cos\phi\sin\theta\hat{\mathbf{x}}+\sin\phi\sin\theta \hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}}) + \dot{\phi}\cos\theta (-\sin\phi\hat{\mathbf{x}}+\cos\phi\hat{\mathbf{y}}) \\ &= - \dot{\theta} \hat{ \mathbf{r}} + \dot{\phi} \cos \theta \hat {\boldsymbol{\phi}} \end{align*} $$
$$ \begin{equation} \begin{aligned} \dot {\hat{ \boldsymbol{\phi}}} &= \frac{d}{dt}(-\sin\phi\hat{\mathbf{x}}+\cos\phi\hat{\mathbf{y}}) \\ &= -\dot\phi\cos\phi\hat{\mathbf{x}}-\dot\phi\sin\phi\hat{\mathbf{y}} \\ &= -\dot\phi (\cos\phi\hat{\mathbf{x}} + \sin\phi\hat{\mathbf{y}}) \end{aligned} \end{equation} $$
The result of $\dot {\hat{\boldsymbol{\phi}}}$, unlike other components, doesn’t straightaway get simplified into $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\theta}}$. If you take a closer look, you’ll notice there’s no $\hat{\mathbf{z}}$ component. By multiplying $\hat{\mathbf{r}}$ by $\sin\theta$ and $\hat{\boldsymbol{\theta}}$ by $\cos\theta$, and then adding them together, the $\hat{\mathbf{z}}$ component disappears. Let’s use this.
$$ \begin{align*} \ \sin\theta\hat{\mathbf{r}}+\cos\theta \hat{\boldsymbol{\theta}} &= \cos\phi\sin^2\theta\hat{\mathbf{x}}+\sin\phi\sin^2\theta\hat{\mathbf{y}}+\sin\theta\cos\theta \hat{\mathbf{z}} \\ &\quad + \cos\phi\cos^2\theta\hat{\mathbf{x}} +\sin\phi\cos^2\theta\hat{\mathbf{y}} -\sin\theta\cos\theta \hat{\mathbf{z}} \\ &= \cos\phi (\sin^2\theta+\cos^2\theta)\hat{\mathbf{x}}+\sin\phi (\sin^2\theta+\cos^2\theta)\hat{\mathbf{y}} \\ &= \cos\phi\hat{\mathbf{x}}+\sin\phi\hat{\mathbf{y}} \end{align*} $$
Substitute the above equation into $(1)$, and it becomes as follows.
$$ \dot{\hat{\boldsymbol{\phi}}}=-\dot\phi\sin\theta\hat{\mathbf{r}} - \dot\phi\cos\theta \hat{\boldsymbol{\theta}} $$
Velocity
Differentiate $\mathbf{r}$ with respect to $t$ to get the following.
$$ \begin{align*} \mathbf{v} = \frac{d \mathbf{r}}{dt} = \frac{d}{dt}(r \hat{\mathbf{r}}) = \frac{d r}{dt}\hat{\mathbf{r}} + r\frac{d \hat{\mathbf{r}}}{dt} &= \dot{r} \hat{\mathbf{r}} + r \dot{\hat{\mathbf{r}}} \\ &= \dot{r} \hat{\mathbf{r}} + r\dot{\theta} \hat{ \boldsymbol{\theta} } + r\dot{\phi} \sin\theta \hat{ \boldsymbol{\phi}} \end{align*} $$
You only need to substitute the precomputed differentiation of the unit vector to get the result.
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Acceleration
Differentiate $\mathbf{v}$ with respect to $t$ to get the following.
$$ \begin{align*} \mathbf{a} = \frac{d \mathbf{v}}{dt} &= \frac{d}{dt}(\dot{r} \hat{\mathbf{r}} +r\dot\theta\hat{\boldsymbol{\theta}}+ r\dot\phi\sin\theta \hat{\boldsymbol{\phi}}) \\ &= (\ddot r \hat{\mathbf{r}} + \dot{r} \dot{\hat{\mathbf{r}}}) + (\dot{r}\dot\theta\hat{\boldsymbol{\theta}} + r\ddot\theta\hat{\boldsymbol{\theta}} + r\dot\theta\dot{\hat{\boldsymbol{\theta}}}) \\ &\quad + (\dot{r} \dot\phi\sin\theta \hat{\boldsymbol{\phi}}+ r\ddot\phi\sin\theta\hat{\boldsymbol{\phi}}+ r\dot\phi\dot\theta\cos\theta\hat{\boldsymbol{\phi}}+ r\dot\phi\sin\theta\dot{\hat{\boldsymbol{\phi}}}) \end{align*} $$
It’s incredibly lengthy. Let’s solve it bit by bit. Since we’ve already computed the differentiation of the unit vector, substituting and simplifying will do the job.
$$ \begin{align*} \mathbf{a} &= \left[ \ddot r \hat{\mathbf{r}} + \dot{r} (\dot{\theta} \hat{\boldsymbol{\theta}} + \dot{\phi} \sin\theta \hat{\boldsymbol{\phi}}) \right] + \left[ \dot{r}\dot\theta\hat{\boldsymbol{\theta}}+ r\ddot\theta\hat{\boldsymbol{\theta}}+r\dot\theta ( -\dot{\theta} \hat{\mathbf{r}} + \dot{\phi} \cos \theta \hat{\boldsymbol{\phi}}) \right] \\ &\quad + \left[ \dot{r} \dot\phi\sin\theta\hat{\boldsymbol{\phi}}+ r\ddot\phi\sin\theta\hat{\boldsymbol{\phi}}+ r\dot\phi\dot\theta\cos\theta\hat{\boldsymbol{\phi}}+ r\dot\phi\sin\theta (-\dot\phi\sin\theta\hat{\mathbf{r}} - \dot\phi\cos\theta\hat{\boldsymbol{\theta}}) \right] \\ &= (\ddot{r}-r\dot\theta^2-r\dot\phi^2\sin^2\theta)\hat{\mathbf{r}}+(r\ddot\theta+2\dot{r}\dot\theta-r\dot\phi^2\sin\theta\cos\theta)\hat{\boldsymbol{\theta}} \\ &\quad +(r\ddot\phi\sin\theta+2\dot{r}\dot\phi\sin\theta+2r\dot\theta\dot\phi\cos\theta)\hat{\boldsymbol{\phi}} \end{align*} $$
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