Solved Problem on Kinematics
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Obtain the speed and acceleration of a body moving in a plane in polar coordinates.


Problem diagram:

In polar coordinates, a point on a trajectory is described by a vector position r making an angle θ with the x-axis. At the considered point, the direction of the position vector, and the variation of the angle are given by the unit vectors êr and êθ.
Figure 1

Solution

The position vector in polar coordinates is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\mathbf{r}=r\;{\mathbf{ê}}_{r}} \tag{I} \end{gather} \]
The velocity vector is given by the derivative with respect to time
\[ \bbox[#99CCFF,10px] {\mathbf{v}=\frac{d\mathbf{r}}{dt}} \]
When the body moves over the trajectory in a time interval, the unit vectors in polar coordinates also change position. This means these vectors are not constant in space (their direction changes point-to-point in the trajectory - Figure 2).
Figure 2

Differentiation of    \( r\;{\mathbf{ê}}_{r} \)

Differentiating the product of functions in the form
\[ (uv)' = u' v+u v' \]
where   \( u=r \)   and   \( v={\mathbf{ê}}_{r} \),   \( u'=\dfrac{dr}{dt} \)   and   \( v'=\dfrac{d{\mathbf{ê}}_{r}}{dt} \)
\[ \frac{d(r\;{\mathbf{ê}}_{r})}{dt} = \frac{dr}{dt}\;{\mathbf{ê}}_{r}+r\;\frac{d{\mathbf{ê}}_{r}}{dt} \]
\[ \mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{dr}{dt}\;{\mathbf{ê}}_{r}+r\;\frac{d{\mathbf{ê}}_{r}}{dt} \]
Interpretation of the terms    \( \dfrac{d{\mathbf{ê}}_{r}}{dt} \)   and   \( \dfrac{d{\mathbf{ê}}_{\theta}}{dt} \)

1stº method

When the body undergoes an infinitesimal displacement dθ, the unit vector in the radial direction moves from êr1 to êr2, and the infinitesimal displacement vector will be (Figure 3)
\[ d{\mathbf{ê}}_{r}={\mathbf{ê}}_{r2}-{\mathbf{ê}}_{r1} \]
this vector will be perpendicular to the unit vector êr, and therefore will be in the same direction that the unit vector êθ
\[ d{\mathbf{ê}}_{r}=d\theta\;{\mathbf{ê}}_{\theta} \]
dividing both sides of the equation by dt
\[ \begin{gather} \frac{d{\mathbf{ê}}_{r}}{dt}=\frac{d\theta}{dt}{\mathbf{ê}}_{\theta} \tag{II} \end{gather} \]

Figure 3: In the figure, the vector displacement dêr does not look perfectly parallel to the unit vector is êθ, this because dθ is exaggerated, if the angle is infinitesimally small, the vectors êr1 and êr2 are close, and dêr is parallel to êθ.

When the body undergoes an infinitesimal displacement dθ the unit vector in the angular direction moves from êθ1 to êθ2, and the infinitesimal displacement vector will be (Figure 4)
\[ d{\mathbf{ê}}_{\theta}={\mathbf{ê}}_{\theta 2}-{\mathbf{ê}}_{\theta1}  \]
this vector will be perpendicular to the unit vector êθ, and therefore will be in the opposite direction to the unit vector êr
\[ d{\mathbf{ê}}_{\theta}=-d\theta\;{\mathbf{ê}}_{r} \]
dividing both sides of the equation by dt
\[ \begin{gather} \frac{d{\mathbf{ê}}_{\theta}}{dt}=-\frac{{d\theta}}{dt}\;{\mathbf{ê}}_{r} \tag{III} \end{gather} \]

Figure 4: In the figure, the vector displacement dêθ does not look perfectly parallel to the unit vector êr, this because dθ is exaggerated, if the angle is infinitesimally small, the vectors êθ1 and êθ2 are close and dêθ it is parallel to êr and pointing to the opposite direction.

2nd º nethod

The position of the unit vectors êr and êθ change with the value of the angle θ and the angle changes over time, so êr and êθ are composite functions of the type
\[ {\mathbf{e}}_{r}[\theta(t)] \qquad \text{e} \qquad {\mathbf{e}}_{\theta}[\theta(t)] \]
The derivatives are given by the chain rule
\[ \begin{gather} \frac{d{\mathbf{e}}_{r}[\theta(t)]}{dt}=\frac{d{\mathbf{e}}_{r}}{d\theta}\frac{d\theta}{dt} \qquad \text{e} \qquad \frac{d{\mathbf{e}}_{\theta}[\theta(t)]}{dt}=\frac{d{\mathbf{e}}_{r}}{d\theta}\frac{d\theta}{dt} \tag{IV} \end{gather} \]
If we decompose the unit vectors êr and êθ from the polar coordinate system in directions i and j of the cartesian coordinate system, we can write (Figure 5)
\[ \begin{gather} {\mathbf{ê}}_{r}={\mathbf{ê}}_{rx}+{\mathbf{ê}}_{ry}\\ {\mathbf{ê}}_{\theta}={\mathbf{ê}}_{\theta x}+{\mathbf{ê}}_{\theta y} \end{gather} \]
\[ \begin{gather} {\mathbf{ê}}_{r}=\cos \theta\;\mathbf{i}+\sin \theta\;\mathbf{j} \tag{V}\\ {\mathbf{ê}}_{\theta}=-\sin \theta\;\mathbf{i}+\cos \theta\;\mathbf{j} \tag{VI} \end{gather} \]

Figure 5

Differentiating the expression (V) with respect to θ
\[ \frac{d{\mathbf{ê}}_{r}}{d\theta}=-\sin \theta\;\mathbf{i}+\cos \theta\;\mathbf{j}={\mathbf{ê}}_{\theta} \]
substituting this value in the first of the expressions (IV)
\[ \frac{d{\mathbf{ê}}_{r}}{dt}=\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta} \]
which agrees with the expression (II) above.
Differentiating the expression (VI) with respect to θ
\[ \frac{d{\mathbf{ê}}_{\theta}}{d\theta}=-\cos \theta\;\mathbf{i}+(-\sin \theta\;\mathbf{j})=-\cos \theta\;\mathbf{i}-\sin \theta\;\mathbf{j}=-(\cos \theta\;\mathbf{i}+\sin \theta\;\mathbf{j})=-{\mathbf{ê}}_{r} \]
substituting this value in the second of expressions (IV)
\[ \frac{d{\mathbf{ê}}_{\theta}}{dt}=-{\frac{d\theta}{dt}}\;{\mathbf{ê}}_{r} \]
which agrees with the expression (III) above.

3rd .º method

Calculating the dot product
\[ {\mathbf{ê}}_{r}.{\mathbf{ê}}_{r}=\left|\;{\mathbf{ê}}_{r}\;\right|.\left|\;{\mathbf{ê}}_{r}\;\right|.\sin 0=1 \]
differentiating this expression with respect to θ
\[ \begin{gather} \frac{d{\mathbf{ê}}_{r}}{d\theta}.{\mathbf{ê}}_{r}+{\mathbf{ê}}_{r}.\frac{d{\mathbf{ê}}_{r}}{d\theta}=\frac{d(1)}{d\theta}\\ 2\;{\mathbf{ê}}_{r}.\frac{d{\mathbf{ê}}_{r}}{d\theta}=0\\ {\mathbf{ê}}_{r}.\frac{d{\mathbf{ê}}_{r}}{d\theta}=0 \end{gather} \]
for this dot product scalar to be equal to zero the vector \( \left(\frac{d{\mathbf{ê}}_{r}}{d\theta}\right) \), it should be perpendicular to the vector êr, remembering that \( \sin \frac{\pi}{2}=0 \), therefore it is in the same direction as the vector êθ, and by Figure 3 it is positive
\[ \frac{d{\mathbf{ê}}_{r}}{d\theta}={\mathbf{ê}}_{\theta} \]
substituting this value in the first of the expressions (IV)
\[ \frac{d{\mathbf{ê}}_{r}}{dt}=\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta} \]
which agrees with the expression (II) above.

Calculating the dot product
\[ {\mathbf{ê}}_{\theta}.{\mathbf{ê}}_{\theta}=\left|\;{\mathbf{ê}}_{\theta}\;\right|.\left|\;{\mathbf{ê}}_{\;\theta}\;\right|.\sin 0=1 \]
differentiating this expression with respect to θ
\[ \begin{gather} \frac{d{\mathbf{ê}}_{\theta}}{d\theta}.{\mathbf{ê}}_{\theta}+{\mathbf{ê}}_{\theta}.\frac{d{\mathbf{ê}}_{\theta}}{d\theta}=\frac{d(1)}{d\theta}\\ 2\;{\mathbf{ê}}_{\theta}.\frac{d{\mathbf{ê}}_{\theta}}{d\theta}=0\\ {\mathbf{ê}}_{\theta}.\frac{d{\mathbf{ê}}_{\theta}}{d\theta}=0 \end{gather} \]
for this dot product scalar to be equal to zero the vector \( \left(\frac{d{\mathbf{ê}}_{\theta}}{d\theta}\right) \), it should be perpendicular to the vector êθ, remembering that \( \sin \frac{\pi}{2}=0 \), therefore it is in the same direction as the vector êr, and by Figure 3 it is negative
\[ \frac{d{\mathbf{ê}}_{\theta}}{d\theta}=-{\mathbf{ê}}_{r} \]
substituting this value in the second of expressions (IV)
\[ \frac{d{\mathbf{ê}}_{\theta}}{dt}=-{\frac{d\theta}{dt}}\;{\mathbf{ê}}_{r} \]
which agrees with the expression (III) above.

using the interpretation made in (II)
\[ \bbox[#FFCCCC,10px] {\mathbf{v}=\frac{dr}{dt}\;\mathbf{ê}_{r}+r\frac{d\theta}{dt}{\mathbf{ê}}_{\theta}} \]
The vector acceleration is given by the derivative with respect to time of the expression above.
\[ \mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d}{dt}\left(\frac{dr}{dt}\;{\mathbf{ê}}_{r}+r\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}\right) \]
Differentiation of    \( \dfrac{dr}{dt}\;\mathbf{ê}_{r} \)

using the product rule:
\[ \frac{d}{dt}\left(\frac{dr}{dt}\;{\mathbf{ê}}_{r}\right)=\frac{d}{dt}\left(\frac{dr}{dt}\right)\;{\mathbf{ê}}_{r}+\frac{dr}{dt}\frac{d\mathbf{ê}_{r}}{dt} \]
using the interpretation made in (II)
\[ \frac{d}{dt}\left(\frac{dr}{dt}\;{\mathbf{ê}}_{r}\right)=\frac{d^{2}r}{dt^{2}}\;{\mathbf{ê}}_{r}+\frac{dr}{dt}\frac{d\theta}{dt}\;{\mathbf{ê}}_{\;\theta} \]

Differentiation of    \( r\dfrac{d\theta}{dt}\;{\mathbf{ê}}_{\theta} \)

differentiating the product of functions in the form
\[ (uvw)' = u' v w+uv'w+u v w' \]
\[ \frac{d}{dt}\left(r\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}\right)=\frac{dr}{dt}\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}+r\frac{d}{dt}\left(\frac{d\theta}{dt}\right)\;{\mathbf{ê}}_{\theta}+r\frac{d\theta}{dt}\frac{d{\mathbf{ê}}_{\theta}}{dt} \]
using the interpretation made in (III)
\[ \begin{gather} \frac{d}{dt}\left(r\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}\right)=\frac{dr}{dt}\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}+r\frac{d^{2}\theta}{dt^{2}}\;{\mathbf{ê}}_{\theta}-r\frac{d\theta}{dt}\frac{d\theta}{dt}\;{\mathbf{ê}}_{r}\\ \frac{d}{dt}\left(r\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}\right)=\frac{dr}{dt}\frac{d\theta}{dt}\;{\mathbf{ê}}_{\theta}+r\frac{d^{2}\theta}{dt^{2}}\;{\mathbf{ê}}_{\theta}-r\left(\frac{d\theta}{dt}\right)^{2}{\mathbf{ê}}_{r} \end{gather} \]
Note: Do not confuse   \( \frac{d}{dt}\left(\frac{d\theta}{dt}\right)=\frac{d^{2}\theta}{dt^{2}} \)   with   \( \frac{d\theta}{dt}\frac{d\theta}{dt}=\left(\frac{d\theta}{dt}\right)^{2} \),   in the first case the second derivative of the angular displacement with respect to time represents the angular acceleration (γ), in the second case the first derivative of the angular displacement with respect to time represents the angular velocity (ω) square, it is the angular velocity squared (ω2).
\[ \begin{gather} \mathbf{a}=\frac{d^{2}r}{dt^{2}}\;{\mathbf{ê}}_{r}+\frac{dr}{dt}\;\frac{d\theta}{dt}{\mathbf{ê}}_{\theta}+\frac{dr}{dt}\frac{d\theta}{dt}{\mathbf{ê}}_{\theta}+r\frac{d^{2}\theta}{dt^{2}}{\mathbf{ê}}_{\theta}-r\left(\frac{d\theta}{dt}\right)^{2}{\mathbf{ê}}_{r}\\ \mathbf{a}=\frac{d^{2}r}{dt^{2}}\;{\mathbf{ê}}_{r}+2\frac{dr}{dt}\;\frac{d\theta}{dt}{\mathbf{ê}}_{\theta}+r\frac{d^{2}\theta}{dt^{2}}\;{\mathbf{ê}}_{\theta}-r\left(\frac{d\theta}{dt}\right)^{2}{\mathbf{ê}}_{r} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\mathbf{a}=\left[\frac{d^{2}r}{dt^{2}}-r\left(\frac{d\theta}{dt}\right)^{2}\right]\;{\mathbf{ê}}_{r}+\left[2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^{2}\theta}{dt^{2}}\right]\;{\mathbf{ê}}_{\theta}} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .