02-Apr-2022

# A background story on how to use Newton’s Law for rotation

This trick is actually quite common. I wish I knew it before learning Newtonian Dynamics in rotations.

A rotation dynamic usually described using vectors. At first I find it quite awkward and foreign because it doesn’t follow exactly how Newton’s law in cartesian coordinate is being used. A simple example is how the Newton’s Law had to be modified using a vector cross product to describe the torque.

$\bm{\tau} =\bm r \times \bm F \\ \sum{\bm \tau} = I \alpha$

This definition works really well for rigid bodies, but turns out very weird in my opinion when dealing with rotation under gravity. In gravity’s case, the rigid bodies can vary in the distance between the axis and itself. Then, the Newton’s equation of motion suddenly contains terms like centripetal acceleration, angular acceleration, and so on.

My high school textbook explains that those acceleration came out due to how polar coordinates behaves differently when we described derivatives.

In the usual x-y cartesian coordinates, we define x or y velocity as the limit of average speed of movement in position.

$\overline{v}_x = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$

In the polar coordinates however, we define the position using a length unit $r$ and a dimensionless unit angle $\theta$, usually in radian. So, if position is described as $(x,y)$ in a cartesian coordinates, it is described as $(r, \theta)$ in polar coordinates.

The weird thing is the direction of the derivatives. For velocity, we have two components as usual.

$\overline{v}_r = \lim_{\Delta t \to 0} \frac{\Delta r}{\Delta t} \\ \overline{v}_\theta = \lim_{\Delta t \to 0} \overline{r}\frac{\Delta \theta}{\Delta t}$

The radial speed is pretty much like the definition of speed in cartesian. But the tangential speed, the speed perpendicular to radial velocity, has somewhat different expression. It relies on an average radial distance, and also depends on the changes in angular position. It is not so obvious to me at that time. However, the textbook justifies the definition using geometrical reasoning.

The geometrical reasoning relates the use of a circle. In a carousel, if two objects is in a different radial position, but both have the same angular speed, both will experience different tangential speed. It is easy imagine that if the radial distance is zero (the object is in the center), the object doesn’t have tangential velocity. So, it is quite easy to understand that the tangential speed scales linearly with the radial distance.

However, this is quite the departure from the usual x and y velocity where both are described the same. One can only describe or derive it by relying on the geometry of circle.

To proceed even further. The textbook describes that in rotation we have 4 kinds of acceleration

$\overline{a}_r = \lim_{\Delta t \to 0} \frac{\Delta v_r}{\Delta t}$

The centripetal acceleration (also in the radial direction, but has to be negative)

$\overline{a}_c = - \frac{v_r^2}{r}$

The coriolis acceleration (in the tangential direction)

$\overline{a}_{cr} = 2 v_r \frac{v_\theta}{r}$

And finally, tangential acceleration (obviously, in the tangential direction as well)

$\overline{a}_{\theta} = \lim_{\Delta t \to 0} \overline{r}\frac{\Delta v_\theta}{\Delta t}$

As you can see, each one of them seems completely different. Only radial and tangential acceleration uses limit. The other seems to be some combination of velocity. Of course, some textbook will use $\omega$ or $\alpha$ as a replacement for angular velocity and acceleration as opposed to tangential term. But the idea is equivalent.

If you relate it with the definition of torque. Things got really interesting because all the acceleration that requires a change in radial position will vanish. So, in torque, you only need to care about the angular acceleration. This is an immediate consequence if you think that rigid bodies will have no change of radial distance.

However, in certain circumstances, like earth rotation in its axis, Coriolis force will have effect. In celestial bodies, you also need to take into account the radial acceleration.

While using vectors, dot product, and cross product is enough to describe rotation. At that time, I feel it is not “general” enough. Even though we know it is derived from Newton’s Law as well. It’s just that the textbook explanation of describing why circle geometry or polar coordinates causes multiple terms of acceleration feels like a forced things to remember, rather than things that you can understand naturally.

# Using complex numbers to describe rotations

Hence, this is why complex numbers is very convenient to describe rotation.

In some cases, complex numbers feels redundant with vectors. The real component basically the x direction, then the imaginary part is basically the y direction.

However, in the case of rotation, it is very convenient because you can express the polar coordinates $(r, \theta)$ as a number. By number, I mean you can add, multiply, or even divide it. In vectors, you can’t divide vectors by vectors because it’s not defined.

By a pretty standard relationship using the exponential function, for any complex number $z$, we have this relationship.

$z = x + i y = r e^{i\theta}$

Where it is defined that:

$r^2 = x^2 + y^2 \\ \tan \theta = \frac{y}{x}$

In the usual usage of Newton’s Law in x-y cartesian coordinates, we usually have two equation (Newton’s Law applied to each direction). In the complex plane, since $z$ is already a number. You can just apply mechanical law to it.

For example, to figure out the velocity, we take the derivative of position with respect to time. I’m going to use the dot notation for this. Placing dot on top of the variable means it is a derivative wrt time (velocity). Two dots means it is derived twice (acceleration). Since position immediately represented as a single complex number, the velocity is then:

$z = r e^{i\theta} \\ \dot{z} = \dot{r}e^{i\theta} + i \dot{\theta} r e^{i\theta} \\ \dot{z} = (\dot{r} + i \dot{\theta} r )e^{i\theta}$

The velocity of the object can be seen immediately that it has two distinct components. The real part $\dot{r}$ and the imaginary part $i\dot{\theta}r$. Both of them are multiplied by the angle argument $e^{i\theta}$. That means the actual projection of the real and imaginary part depends on the angle. However, it is easily seen that magnitude of the velocity contains radial direction $\dot{r}$ (because it is real) and tangential direction $i\dot{\theta}r$ (because it has imaginary number).

If we take the derivative again wrt time, we got the acceleration.

$\dot{z} = \dot{r}e^{i\theta} + i \dot{\theta} r e^{i\theta} \\ \ddot{z} = \ddot{r}e^{i\theta}+ i\dot{r}\dot{\theta}e^{i\theta} + i\ddot{\theta}r e^{i\theta} + i\dot{\theta}\dot{r}e^{i\theta} + i i r \dot{\theta}^2 e^{i\theta} \\ \ddot{z} = [(\ddot{r}-r\dot{\theta}^2)+i(2\dot{r}\dot{\theta}+\ddot{\theta}r)]e^{i\theta}$

Same as before, you will recognize there are 4 terms neatly organized in the real and imaginary part. The real part is in the radial direction, which is a combination of two terms:

$\ddot{r}-r\dot{\theta}^2$

The imaginary part is in the tangential direction, which is a combination of two terms:

$i(2\dot{r}\dot{\theta}+\ddot{\theta}r)$

You can now easily associate the terms with it’s respective definition of acceleration:

$a_r = \ddot{r} \\ a_c = - r \dot{\theta}^2 \\ a_{cr} = 2 \dot{r}\dot{\theta} \\ a_\theta = r\ddot{\theta}$

Next time, if you forget it, you can just derive it using algebra.

# Using Newton’s Law in complex plane

Now that we know position, velocity, and acceleration in complex numbers, how do we use it in Newton’s Law?

Like the position, you also describe force or torque as complex numbers.

Consider a rigid body with applied torque at a given radial distance $L$ and angle $\phi$.

The torque is just the force (as complex numbers) multiplied by the position of the force (as complex numbers).

$\tau = Le^{i\theta}F e^{i\phi} = FLe^{i(\theta+\phi)}$

For the acceleration terms, note that in rigid bodies the radial distance of the object’s particles will not change. Hence it will only contains the rotation inertial quantity and any acceleration with constant $r$. Which is angular and centripetal acceleration. However, notice that the complex numbers of the angular acceleration is its magnitude times imaginary unit, times the angle argument (because it is in the tangential direction). While, the complex numbers of centripetal acceleration is its magnitude times the angle argument (because it is in the radial direction).

$\tau = I(-\dot{\theta}^2 + i\ddot{\theta}) e^{i\theta} \\ FLe^{i(\theta+\phi)} = I(-\dot{\theta}^2 + i\ddot{\theta}) e^{i\theta} \\ FLe^{i\phi} = I(-\dot{\theta}^2 + i\ddot{\theta})$

It will immediately follow that the right hand side contains both real and imaginary components. That means, it depends on the angle $\phi$. If $\phi=0$ that means the left hand side is entirely real. So the bodies will have instantaneous centripetal acceleration along its axis. But it will have zero angular acceleration in the imaginary part. On the other hand. If $\phi=\frac{\pi}{2}$, it will have no real component, but it will have imaginary component. In which the bodies instantaneously undergoes angular acceleration.

It also works the same way if $\phi$ (the angle between the force and it’s position in the polar coordinate) is not purely real or imaginary, but as an angle in between.

This looks really complex, but usually rigid rotation problems involves fixed axis, so we can just concern ourselves with angular/tangential direction, which is the imaginary part.

Because of this, complex numbers seems a little bit of overkill to be used for rigid body rotations. Most of the time, we are only dealing with the imaginary part.

Despite that, complex numbers provides a very helpful mnemonic on why we have many terms of acceleration for rotations. It also provides an insight on why torque is defined by a cross product of position vector and force vector. It is essentialy analoguous with complex number multiplication, but we only want to retrieve the imaginary part. It is just that in vector terms, we can’t just multiply by angles like in the complex numbers. So we multiply it with position vector to get its direction.

# Remark

We conclude the article for today. Next, we will use complex numbers to describe rotation under gravity, which is a more exciting problem.

Written by Rizky Maulana Nugraha
Software Developer. Currently remotely working from Indonesia.