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Velocity and Acceleration in Polar Coordinates 📂Classical Mechanics

Velocity and Acceleration in Polar Coordinates

Velocity and Acceleration in Polar Coordinates

v=r˙r^+rθ˙θ^a=(r¨−rθ˙2)r^+(2r˙θ˙+rθ¨)θ^ \begin{align*} \mathbf{v}&=\dot{r} \hat{\mathbf{r}} + r \dot{\theta} \hat{\boldsymbol{\theta}} \\ \mathbf{a}&= (\ddot r -r\dot{\theta} ^2)\hat{\mathbf{r}} + (2\dot{r} \dot{\theta} + r\ddot{\theta})\hat{\boldsymbol{\theta}} \end{align*}

Derivation

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In the polar coordinate system, unit vectors can be described as follows.

r=rr^=xx^+yy^  ⟹  r^=xrx^+yry^=cos⁡θx^+sin⁡θy^=r^(θ)θ^=r^(θ+π/2)=−sin⁡θx^+cos⁡θy^ \begin{align*} && \mathbf{r}&=r\hat{\mathbf{r}}=x\hat{\mathbf{x}} + y \hat{\mathbf{y}} \\ \implies && \hat{\mathbf{r}} &= \frac{x}{r}\hat{\mathbf{x}} +\frac{y}{r} \hat{\mathbf{y}}=\cos\theta \hat{\mathbf{x}} + \sin\theta \hat{\mathbf{y}} = \hat{\mathbf{r}} (\theta) \\ {} \\ && \hat \theta &= \hat{\mathbf{r}}(\theta+\pi/2)= -\sin\theta \hat{\mathbf{x}} + \cos\theta \hat{\mathbf{y}} \end{align*}

Velocity is obtained by differentiating position with respect to time, and acceleration is obtained by differentiating velocity with respect to time. Note that r˙\dot{r} is pronounced “dot”. In physics, a dot above a letter means differentiation with respect to time.

r˙=drdt \dot{r}=\frac{dr}{dt}

Velocity

Differentiating r\mathbf{r} with respect to tt gives the following.

v=drdt=ddt(rr^)=drdtr^+rdr^dt=r˙r^+rr^˙ \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}}}

Calculating r^˙\dot{\hat{\mathbf{r}}}, since x^\hat{\mathbf{x}} and y^\hat{\mathbf{y}} do not change over time, dx^dt=0\dfrac{d \hat{\mathbf{x}}}{dt}=0 is true. Hence, it can be given as follows.

r^˙=ddt(r^)=ddt(cos⁡θx^)+ddt(sin⁡θy^)=cos⁡θdtx^+sin⁡θdty^=cos⁡θdθdθdtx^+sin⁡θdθdθdty^=−sin⁡θdθdtx^+cos⁡θdθdty^=dθdt(−sin⁡θx^+cos⁡θy^)=θ˙θ^ \begin{align*} \dot{\hat{\mathbf{r}}} = \frac{d}{dt}(\hat{\mathbf{r}}) &= \frac{d}{dt}(\cos\theta \hat{\mathbf{x}}) + \frac{d}{dt}(\sin\theta \hat{\mathbf{y}}) \\ &= \frac{\cos\theta}{dt}\hat{\mathbf{x}} + \frac{\sin\theta}{dt}\hat{\mathbf{y}} \\ &= \frac{\cos\theta}{d \theta}\frac{d \theta}{dt}\hat{\mathbf{x}}+\frac{\sin\theta}{d \theta}\frac{d \theta}{dt}\hat{\mathbf{y}} \\ &= -\sin\theta \frac{d \theta}{dt}\hat{\mathbf{x}}+\cos\theta \frac{d \theta}{dt}\hat{\mathbf{y}} \\ &= \frac{d \theta }{dt}(-\sin\theta \hat{\mathbf{x}}+\cos\theta \hat{\mathbf{y}}) \\ &= \dot{\theta} \hat{\boldsymbol{\theta}} \end{align*} Therefore, the velocity in polar coordinates is as follows.

v=r˙r^+rθ˙θ^ \mathbf{v}=\dot{r} \hat{\mathbf{r}} + r \dot{\theta} \hat{\boldsymbol{\theta}}

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Acceleration

Differentiating v\mathbf{v} with respect to tt gives the following.

a=dvdt=d(r˙r^+rθ˙θ^)dt=r¨r^+r˙r^˙+r˙θ˙θ^+rθ¨θ^+rθ˙θ^˙ \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \dfrac{d(\dot{r} \hat{\mathbf{r}} + r \dot{\theta} \hat{\boldsymbol{\theta}})}{dt} = \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}}}

Calculating θ^˙\dot{ \hat{\boldsymbol{\theta}}} gives the following result.

θ^˙=ddt(θ^)=ddt(−sin⁡θx^)+ddt(cos⁡θy^)=−dsin⁡θdtx^+dcos⁡θdty^=−dsin⁡θdθdθdtx^+dcos⁡θdθdθdty^=dθdt(−cos⁡θx^−sin⁡θy^)=−θ˙r^ \begin{align*} \dot{ \hat{\boldsymbol{\theta}}} = \frac{d}{dt}(\hat{\boldsymbol{\theta}}) &= \frac{d}{dt}(-\sin\theta \hat{\mathbf{x}})+\frac{d}{dt}(\cos\theta \hat{\mathbf{y}}) \\ &= -\frac{d \sin\theta}{dt}\hat{\mathbf{x}} +\frac{d\cos\theta}{dt}\hat{\mathbf{y}} \\ &= -\frac{d\sin\theta}{d \theta}\frac{d \theta}{dt}\hat{\mathbf{x}}+\frac{d\cos\theta}{d \theta}\frac{d \theta}{dt}\hat{\mathbf{y}} \\ &= \dfrac{d\theta}{dt} (-\cos\theta \hat{\mathbf{x}}-\sin\theta \hat{\mathbf{y}}) \\ &= - \dot{\theta} \hat{\mathbf{r}} \end{align*}

Since r^˙\dot{\hat{\mathbf{r}}} was calculated when deriving velocity, substituting and organizing gives the following result.

a=r¨r^+r˙r^˙+r˙θ˙θ^+rθ¨θ^+rθ˙θ^˙=r¨r^+r˙θ˙θ^+r˙θ˙θ^+rθ¨θ^−rθ˙θ˙r^=(r¨−rθ˙2)r^+(2r˙θ˙+rθ¨)θ^ \begin{align*} \mathbf{a} &= \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}}} \\ &= \ddot r \hat{\mathbf{r}} +\dot{r} \dot{\theta} \hat{\theta} + \dot{r} \dot{\theta} \hat{\boldsymbol{\theta}} + r \ddot{\theta} \hat{\boldsymbol{\theta}} -r \dot{\theta} \dot{\theta} \hat{\mathbf{r}} \\ &= (\ddot r -r\dot{\theta} ^2)\hat{\mathbf{r}} + (2\dot{r} \dot{\theta} + r\ddot{\theta})\hat{\boldsymbol{\theta}} \end{align*}

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