logo

Velocity and Acceleration in Cylindrical Coordinates 📂Classical Mechanics

Velocity and Acceleration in Cylindrical Coordinates

Velocity and Acceleration in Cylindrical Coordinates

v=r˙r^+rϕ˙ϕ^+z˙z^a=(r¨−rϕ˙2)r^+(2r˙ϕ˙+rϕ¨)ϕ^+z¨z^ \begin{align*} \mathbf{v}&=\dot{r} \hat{\mathbf{r}} + r \dot{\phi} \hat{\boldsymbol{\phi}}+\dot{z} \hat{\mathbf{z}} \\ \mathbf{a} &= (\ddot r -r\dot{\phi} ^2)\hat{\mathbf{r}} + (2\dot{r} \dot{\phi} + r\ddot{\phi})\hat{\boldsymbol{\phi}} + \ddot{z}\hat{\mathbf{z}} \end{align*}

Derivation

596DB6CE0.jpg

In cylindrical coordinates, the unit vectors are as follows.

ρ=xx^+yy^+zz^=rr^+zz^r^=r^(ϕ)=cos⁡ϕx^+sin⁡ϕy^ϕ^=r^(ϕ+π/2)=−sin⁡ϕx^+cos⁡ϕy^z^=z^ \begin{align*} \boldsymbol{\rho}&=x\hat{\mathbf{x}}+y \hat{\mathbf{y}} +z\hat{\mathbf{z}}=r\hat{\mathbf{r}} +z\hat{\mathbf{z}} \\ \hat{\mathbf{r}} &= \hat{\mathbf{r}}(\phi) = \cos\phi \hat{\mathbf{x}} + \sin\phi \hat{\mathbf{y}} \\ \hat{\boldsymbol{\phi}} &= \hat{\mathbf{r}}(\phi+\pi/2) = -\sin\phi \hat{\mathbf{x}} + \cos\phi \hat{\mathbf{y}} \\ \hat{\mathbf{z}} &= \hat{\mathbf{z}} \end{align*} Velocity is obtained by differentiating position with respect to time, and acceleration is obtained by differentiating velocity with respect to time. For reference, r˙\dot{r} is read as “dot”. In physics, a dot over a letter signifies differentiation with respect to time.

r˙=drdt \dot{r} = \frac{dr}{dt}

Velocity

By differentiating ρ\boldsymbol{\rho} with respect to tt, we get the following.

v=dρdt=ddt(rr^+zz^)=drdtr^+rdr^dt+dzdtz^+zdz^dt=r˙r^+rr^˙+z˙z^ \begin{align*} \mathbf{v}&=\frac{d \boldsymbol{\rho}}{dt} \\ &=\frac{d}{dt}(r \hat{\mathbf{r}}+z\hat{\mathbf{z}}) \\ &=\frac{d r}{dt}\hat{\mathbf{r}} + r\frac{d \hat{\mathbf{r}}}{dt} +\frac{dz}{dt}\hat{\mathbf{z}} +z\frac{d \hat{\mathbf{z}}}{dt} \\ &=\dot{r} \hat{\mathbf{r}} +r \dot{\hat {\mathbf{r}}}+\dot{z} \hat{\mathbf{z}} \end{align*}

At this point, unit vectors in Cartesian coordinates do not change over time. In other words, it is as follows.

x^˙=y^˙=z^˙=0 \dot{\hat{\mathbf{x}}}=\dot{\hat{\mathbf{y}}}=\dot{\hat{\mathbf{z}}} = \mathbf{0}

Calculating r^˙\dot{\hat{\mathbf{r}}} gives us the following.

r^˙=ddt(r^)=ddt(cos⁡ϕx^)+ddt(sin⁡ϕy^)=dcos⁡ϕdtx^+dsin⁡ϕdty^=dcos⁡ϕdϕdϕdtx^+dsin⁡ϕdϕdϕdty^=−sin⁡ϕdϕdtx^+cos⁡ϕdϕdty^=dϕdt(−sin⁡ϕx^+cos⁡ϕy^)=ϕ˙ϕ^ \begin{align*} \dot{\hat {\mathbf{r}}} = \frac{d}{dt}(\hat{\mathbf{r}}) &= \frac{d}{dt}(\cos\phi \hat{\mathbf{x}}) + \frac{d}{dt}(\sin\phi \hat{\mathbf{y}}) \\ &= \frac{d\cos\phi}{dt}\hat{\mathbf{x}} + \frac{d\sin\phi}{dt}\hat{\mathbf{y}} \\ &= \frac{d\cos\phi}{d \phi}\frac{d \phi}{dt}\hat{\mathbf{x}}+\frac{d\sin\phi}{d \phi}\frac{d \phi}{dt}\hat{\mathbf{y}} \\ &= -\sin\phi \frac{d \phi}{dt}\hat{\mathbf{x}}+\cos\phi \frac{d \phi}{dt}\hat{\mathbf{y}} \\ &= \frac{d \phi }{dt}(-\sin\phi \hat{\mathbf{x}}+\cos\phi \hat{\mathbf{y}}) \\ &= \dot{\phi} \hat{\boldsymbol{\phi}} \end{align*}

Therefore, the velocity is as follows.

v=r˙r^+rϕ˙ϕ^+z˙z^ \mathbf{v}=\dot{r} \hat{\mathbf{r}} + r \dot{\phi} \hat{\boldsymbol{\phi}}+\dot{z} \hat{\mathbf{z}}

■

Acceleration

By differentiating v\mathbf{v} with respect to tt, we get the following.

a=dvdt=ddt(r˙r^+rϕ˙ϕ^+z˙z^)=r¨r^+r˙r^˙+r˙ϕ˙ϕ^+rϕ¨ϕ^+rϕ˙ϕ^˙+z¨z^ \begin{align*} \mathbf{a}=\frac{d \mathbf{v}}{dt} &= \frac{d}{dt}(\dot{r} \hat{\mathbf{r}} + r \dot{\phi} \hat{\boldsymbol{\phi}} +\dot{z} \hat{\mathbf{z}}) \\ &= \ddot r \hat{\mathbf{r}} +\dot{r} \dot{ \hat{\mathbf{r}}} + \dot{r} \dot{\phi} \hat{\boldsymbol{\phi}} + r \ddot{\phi} \hat{\boldsymbol{\phi}} + r \dot{\phi} \dot{ \hat{\boldsymbol{\phi}}} +\ddot{z}\hat{\mathbf{z}} \end{align*}

Upon calculating ϕ^˙\dot{\hat{\boldsymbol{\phi}}}, we get the following.

ϕ^˙=ddt(ϕ^)=ddt(−sin⁡ϕx^)+ddt(cos⁡ϕy^)=−dsin⁡ϕdtx^+dcos⁡ϕdty^=−dsin⁡ϕdϕdϕdtx^+dcos⁡ϕdϕdϕdty^=dϕdt(−cos⁡ϕx^−sin⁡ϕy^)=−ϕ˙r^ \begin{align*} \dot{ \hat{\boldsymbol{\phi}}} = \frac{d}{dt}(\hat{\boldsymbol{\phi}}) &= \frac{d}{dt}(-\sin\phi \hat{\mathbf{x}})+\frac{d}{dt}(\cos\phi \hat{\mathbf{y}}) \\ &= -\frac{d\sin\phi}{dt}\hat{\mathbf{x}} +\frac{d\cos\phi}{dt}\hat{\mathbf{y}} \\ &= -\frac{d\sin\phi}{d \phi}\frac{d \phi}{dt}\hat{\mathbf{x}}+\frac{d\cos\phi}{d \phi}\frac{d \phi}{dt}\hat{\mathbf{y}} \\ &= \dfrac{d\phi}{dt} (-\cos\phi \hat{\mathbf{x}}-\sin\phi \hat{\mathbf{y}}) \\ &= - \dot{\phi} \hat{\mathbf{r}} \end{align*}

Therefore, substituting and arranging gives us the following.

a=r¨r^+r˙r^˙+r˙ϕ˙ϕ^+rϕ¨ϕ^+rϕ˙ϕ^˙+z¨z^=r¨r^+r˙ϕ˙ϕ^+r˙ϕ˙ϕ^+rϕ¨ϕ^−rϕ˙ϕ˙r^+z¨z^=(r¨−rϕ˙2)r^+(2r˙ϕ˙+rϕ¨)ϕ^+z¨z^ \begin{align*} \mathbf{a} &= \ddot r \hat{\mathbf{r}} +\dot{r} \dot{ \hat{\mathbf{r}}} + \dot{r} \dot{\phi} \hat{\boldsymbol{\phi}} + r \ddot{\phi} \hat{\boldsymbol{\phi}} + r \dot{\phi} \dot{ \hat{\boldsymbol{\phi}}} +\ddot{z}\hat{\mathbf{z}} \\ &= \ddot r \hat{\mathbf{r}} +\dot{r} \dot{\phi}\hat{\boldsymbol{\phi}} + \dot{r} \dot{\phi} \hat{\boldsymbol{\phi}} + r \ddot{\phi} \hat{\boldsymbol{\phi}} -r \dot{\phi} \dot{\phi} \hat{\mathbf{r}} +\ddot{z}\hat{\mathbf{z}} \\ &= (\ddot r -r\dot{\phi} ^2)\hat{\mathbf{r}} + (2\dot{r} \dot{\phi} + r\ddot{\phi})\hat{\boldsymbol{\phi}} +\ddot{z}\hat{\mathbf{z}} \end{align*}

■

Since the cylindrical coordinate system is just the polar coordinate system with the addition of height zz, the formulas for velocity and acceleration also simply add terms z˙z^\dot{z} \hat{\mathbf{z}} and z¨z^\ddot{z}\hat{\mathbf{z}} respectively to their polar coordinate system counterparts.

See Also