# 2D Rotational Dynamics

## Alexei Gilchrist

The description of 2D rotational motion is further developed in analogue to linear dynamics by introducing the rotational equivalents of force, momentum and mass.

### 1 Torque

We want to find the rotational analogue of force. To do this we start by looking at the work done by a force that rotates an object by a small angle $$\Delta\theta$$. The work done will be the force times the distance, which we can split into $$x$$ and $$y$$ components:

\begin{align*}\Delta W &= (\mbox{force})\times(\mbox{distance}) \\ &= F_x \Delta x + F_y \Delta y \\ &= (F_y x - F_x y) \Delta\theta \\ &\sim (\mbox{rotational force})\times(\mbox{rotational distance})\end{align*}
In the 3rd line we made use of $$\Delta x = -y \Delta\theta$$ and $$\Delta y = x \Delta\theta$$ for rotational motion (see 2d-rotational-kinematics). This expression would have the same functional form for the work done as for linear motion if we identify the term in the brackets as the rotational analogue of force—this term is called the torque $$\tau$$
$\begin{equation*}\tau = F_y x - F_x y\end{equation*}$

There are several ways of writing the torque which are related to each other geometrically. Consider the following diagram

A force $$\vec{F}$$ is applied at position $$\vec{r}$$ from the pivot, and it causes a small rotation $$\Delta \theta$$ of the object. The work done by the force is the component of the force in the direction of the motion times the distance traveled. From the diagram, this is $$\Delta W = r \Delta\theta F_\perp$$. We know from the above argument that this must be equal to $$\Delta W = \tau \Delta\theta$$, so we have that $$\tau = r F_\perp$$. Looking at the diagram again, the perpendicular component of the force will be $$F_\perp = F \sin\theta$$. So we have an additional three convenient ways of expressing the torque:

\begin{align*}\tau &= rF\sin\theta \\ &= r (F\sin\theta) = r F_\perp \\ &= (r \sin\theta) F = r_\perp F.\end{align*}
In words, the torque is the product of the magnitudes of the moment arm ($$r$$) and the force, times the sine of the angle between them. This is the same as the product of the moment arm and the perpendicular component of the force; or of the force and the perpendicular component of the moment arm.

Say we had a whole lot of particles that where contributing to the work done, then

$\begin{equation*}\Delta W_\mathrm{tot} = \sum_j (F_{yj}x_j-F_{xj}y_j)\Delta\theta = \sum_j \tau_j \Delta\theta = \tau_\mathrm{tot} \Delta\theta.\end{equation*}$
So with lots of particles, this rotational analogue of a force also adds together into a total torque as the force does in linear motion.

### 2 Angular Momentum

For motion along a line the force is the rate of change of the momentum $$\vec{F}=\dot{\vec{p}}$$. Since we now have the torque as the rotational analogue to force, let’s look for a quantity that plays the same role as the momentum.

$\begin{equation*}\tau = F_y x - F_x y = x (m \ddot{y}) - y (m \ddot{x}).\end{equation*}$
It’s not immediately clear that we could re-write this expression as the time derivative of something. Indeed we can, but it’s easiest to proceed backwards and expand the following:
$\begin{equation*}\frac{d}{dt}(m x \dot{y} - m y \dot{x}) = m\dot{x}\dot{y} + m x \ddot{y} - m \dot{y} \dot{x} -m y \ddot{x} = m x \ddot{y}-m y \ddot{x}.\end{equation*}$
So we have
$\begin{equation*}\tau = \frac{d}{dt} (m x \dot{y} - m y \dot{x}) = \dot{L},\end{equation*}$
where we have identified the angular momentum $$L$$ as
$\begin{equation*}L = m (x \dot{y} - y \dot{x})\end{equation*}$
It has the same relationship to the torque as the momentum has to the force. Since the momentum is $$\vec{p} = m\dot{\vec{r}}$$ we could also write this as
$\begin{equation*}L = x p_y - y p_x.\end{equation*}$
In fact, given the moment arm $$\vec{r}$$ and the momentum $$\vec{p}$$ we can draw a diagram very similar to what was done for the torque and there are a number of useful ways you could express the angular momentum:
\begin{align*}L &= rp\sin\theta \\ &= r p_\perp \\ &= r_\perp p.\end{align*}
where $$\theta$$ is the angle between $$\vec{r}$$ and $$\vec{p}$$.

### 3 Moment of Inertia

Consider some object that is rotating about some pivot. Imagine we split up the object into lots of little masses each at a position $$\vec{r}_j$$ from the pivot and moving with velocity $$\vec{v}_j$$. The total kinetic energy of the object will be

$\begin{equation*}T = \sum_j \frac{1}{2}m_j {v}_j^2\end{equation*}$
where the speed of each piece of mass will be $$v_j=r_j \dot{\theta}_j$$ since the object is rotating. Also each $$\dot{\theta}_j$$ will be the same or else the object will break apart with some pieces rotating faster than others! Labeling this common angular velocity as $$\dot{\theta}$$ we have
$\begin{equation*}T = \frac{1}{2}\left(\sum_j m_jr_j^2\right)\dot{\theta}^2.\end{equation*}$
Now, $$\dot{\theta}$$ is the angular velocity, so if we identified the term in the brackets as the rotational equivalent of mass we would have the same form for the kinetic energy as we do for linear motion: $$T=1/2 m v^2$$, but with rotational equivalents.

The term in the brackets is called the moment of inertia and it indeed does play the role of a mass, quantifying how difficult it is to rotate an object. Normally the moment of inertial is given the symbol $$I$$:

$\begin{equation*}I = \sum_j m_j r_j^2.\end{equation*}$
If the object was a continuous object, the sum becomes an integral over the mass
$\begin{equation*}I = \int_\mathrm{obj} r^2 dm.\end{equation*}$
This integral is actually quite a compact notation for the idea of the moment of inertia. To be able to do a calculation, you need to know how the mass is distributed in the object. If it’s uniformly distributed then the integral will reduce to an integration of $$r^2$$ over the volume of the object multiplied by the density. It takes a little practice to get the hang of unpacking this integral into a form you can compute with.

Finally, The angular momentum for a purely rotating object is $$L_\mathrm{tot} =\sum_j L_j = \sum_j r_j p_j$$ because each $$\vec{r}_j$$ and $$\vec{p}_j$$ will be perpendicular to each other. So we have

$\begin{equation*}L_\mathrm{tot} = \sum_j r_j p_j= \sum_j r_j (m_j v_j)= \sum_j r_j (m_j r_j \dot{\theta}) = \left(\sum_j m_j r_j^2\right) \dot{\theta} = I \dot{\theta}.\end{equation*}$
This is of course the rotational equivalent of $$p=mv$$ so everything hangs together nicely$$\ldots$$

$$\ldots$$ except, well, it doesn’t. This nice analogue we’ve built up breaks down in 3D!