4.2 Angular Kinematics

Similar to how we analysed motion along each axis seperately in cartesian coordinates in chapter 1, we can do the same thing in polar coordinates, except with some differences. In the special case of circular motion, there is no motion in the radial direction and 2D motion is reduced to a 1D problem. Displacement has units of length, but angles are unitless, so we will now examine how the angle of the position vector changes over time.

Definition 4.1.

The angular displacement of an object is the difference in angle to the $x$ axis between two times $t_{1}$ and $t_{2}>t_{1}$ .

\Delta\theta=\theta(t_{2})-\theta(t_{1})=\theta_{2}-\theta_{1}.

(4.13)

Now consider the velocity of the object, analogously to the linear case, we can define the angular velocity as the rate of change of angular displacement.

Definition 4.2.

The instananeous angular velocity of an object is defined as the time derivative of angular displacement.

\omega(t)=\lim_{\Delta t\to 0}\frac{\theta(t+\Delta t)-\theta(t)}{\Delta t}=% \frac{\operatorname{\mathrm{d}}\!\theta(t)}{\operatorname{\mathrm{d}}\!t}=\dot% {\theta}.

(4.14)

Likewise, the angular acceleration is given by the rate of change of angular velocity.

Definition 4.3.

The instaneous angular acceleration of an object is defined as the time derivative of angular velocity.

\alpha(t)=\lim_{\Delta t\to 0}\frac{\omega(t+\Delta t)-\omega(t)}{\Delta t}=% \frac{\operatorname{\mathrm{d}}\!\omega(t)}{\operatorname{\mathrm{d}}\!t}=% \frac{\operatorname{\mathrm{d}}\!^{2}\theta(t)}{\operatorname{\mathrm{d}}\!t^{% 2}}=\dot{\omega}=\ddot{\theta}.

(4.15)

Again analogously to the linear case, we define the average angular velocity and angular acceleration as integrals.

	$\displaystyle\bar{\omega}(t)$	$\displaystyle=\frac{\Delta\theta}{\Delta t}=\frac{1}{\Delta t}\int_{t_{1}}^{t_% {2}}\omega(t)\operatorname{\mathrm{d}}\!t$		(4.16)
	$\displaystyle\bar{\alpha}(t)$	$\displaystyle=\frac{\Delta\omega}{\Delta t}=\frac{1}{\Delta t}\int_{t_{1}}^{t_% {2}}\alpha(t)\operatorname{\mathrm{d}}\!t.$		(4.17)

4.2.1 Motion in Polar Coordinates

Now, let us go back to the full 2D vectors and see how they change with time. The velocity vector is the time derivative of the position vector:

\boldsymbol{v}=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{r}}{\operatorname{% \mathrm{d}}\!t}=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!% t}(r\hat{\boldsymbol{r}}).

(4.18)

We need to use the product rule for this since in general both $r$ (the length of the position vector) and $\hat{\boldsymbol{r}}$ (the direction that the position vector points) change over time. What is the time derivative of $\hat{\boldsymbol{r}}$ ? We can work it out using equation 4.1:

$\displaystyle\frac{\operatorname{\mathrm{d}}\!\hat{\boldsymbol{r}}}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}% \cos\theta\hat{\boldsymbol{\imath}}+\frac{\operatorname{\mathrm{d}}\!}{% \operatorname{\mathrm{d}}\!t}\sin\theta\hat{\boldsymbol{\jmath}}$	(4.19)
	$\displaystyle=-\dot{\theta}\sin\theta\hat{\boldsymbol{\imath}}+\dot{\theta}% \cos\theta\hat{\boldsymbol{\jmath}}$	(4.20)
	$\displaystyle=\omega(-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}})$	(4.21)
	$\displaystyle=\omega\hat{\boldsymbol{\theta}},$	(4.22)

where we have used equation 4.2 to substitute $\hat{\boldsymbol{\theta}}$ in the last line. We can work out the time derivative of $\hat{\boldsymbol{\theta}}$ now as well, it is

$\displaystyle\frac{\operatorname{\mathrm{d}}\!\hat{\boldsymbol{\theta}}}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=-\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t% }\sin\theta\hat{\boldsymbol{\imath}}+\frac{\operatorname{\mathrm{d}}\!}{% \operatorname{\mathrm{d}}\!t}\cos\theta\hat{\boldsymbol{\jmath}}$	(4.23)
	$\displaystyle=-\dot{\theta}\cos\theta\hat{\boldsymbol{\imath}}-\dot{\theta}% \sin\theta\hat{\boldsymbol{\jmath}}$	(4.24)
	$\displaystyle=-\omega\hat{\boldsymbol{r}}.$	(4.25)

As an aside, think about why these two results make physical sense. Why should the rate of change of $\hat{\boldsymbol{r}}$ be in the $\hat{\boldsymbol{\theta}}$ direction?

We can now finish off calculating the velocity vector:

	$\displaystyle\boldsymbol{v}$	$\displaystyle=\dot{r}\hat{\boldsymbol{r}}+r\frac{\operatorname{\mathrm{d}}\!% \hat{\boldsymbol{r}}}{\operatorname{\mathrm{d}}\!t}$		(4.26)
		$\displaystyle=\dot{r}\hat{\boldsymbol{r}}+r\omega\hat{\boldsymbol{\theta}}.$		(4.27)

We label the component pointing along the radial direction $v_{r}=\dot{r}$ and call it the radial velocity, and the component pointing along $\hat{\boldsymbol{\theta}}$ , $v_{\theta}=r\omega$ , is the tangential velocity.

The benefit of using the $\hat{\boldsymbol{r}}-\hat{\boldsymbol{\theta}}$ axes with polar coordinates is immediately apparent. Using the cartesian coordinate axes, the position vector is given by $\boldsymbol{r}=r\cos\theta\hat{\boldsymbol{\imath}}+r\sin\theta\hat{% \boldsymbol{\jmath}}$ , so the time derivative is

$\displaystyle\boldsymbol{v}=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{r}}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=(\dot{r}\cos\theta-r\omega\sin\theta)\hat{\boldsymbol{\imath}}+(% \dot{r}\sin\theta+r\omega\cos\theta)\hat{\boldsymbol{\jmath}}$	(4.28)
	$\displaystyle=\dot{r}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})+r\omega(-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta% \hat{\boldsymbol{\jmath}})$	(4.29)
	$\displaystyle=\dot{r}\hat{\boldsymbol{r}}+r\omega\hat{\boldsymbol{\theta}}.$	(4.30)

In the last two lines we have regrouped the terms to write $\boldsymbol{v}$ in the $\hat{\boldsymbol{r}}-\hat{\boldsymbol{\theta}}$ basis and recovered our previous result. In the cartesian coordinate system, we get $v_{x}=\dot{r}\cos\theta-r\omega\sin\theta$ and $v_{y}=\dot{r}\sin\theta+r\omega\cos\theta$ which is rather opaque to interpretation.

Now, let’s calculate the acceleration vector in polar coordinates.

$\displaystyle\boldsymbol{a}=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{v}}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}% (\dot{r}\hat{\boldsymbol{r}}+r\omega\hat{\boldsymbol{\theta}})$	(4.31)
	$\displaystyle=\ddot{r}\hat{\boldsymbol{r}}+\dot{r}\frac{\operatorname{\mathrm{% d}}\!\hat{\boldsymbol{r}}}{\operatorname{\mathrm{d}}\!t}+\dot{r}\dot{\theta}% \hat{\boldsymbol{\theta}}+r\ddot{\theta}\hat{\boldsymbol{\theta}}+r\dot{\theta% }\frac{\operatorname{\mathrm{d}}\!\hat{\boldsymbol{\theta}}}{\operatorname{% \mathrm{d}}\!t}$	(4.32)
	$\displaystyle=\ddot{r}\hat{\boldsymbol{r}}+\dot{r}\dot{\theta}\hat{\boldsymbol% {\theta}}+\dot{r}\dot{\theta}\hat{\boldsymbol{\theta}}+r\ddot{\theta}\hat{% \boldsymbol{\theta}}-r\dot{\theta}^{2}\hat{\boldsymbol{\theta}}$	(4.33)
	$\displaystyle=(\ddot{r}-r\dot{\theta}^{2})\hat{\boldsymbol{r}}+(2\dot{r}\dot{% \theta}+r\ddot{\theta})\hat{\boldsymbol{\theta}}.$	(4.34)

Therefore, the radial acceleration is $a_{r}=\ddot{r}-r\dot{\theta}^{2}$ , and the tangential acceleration is $a_{\theta}=2\dot{r}\dot{\theta}+r\ddot{\theta}$ .

Here is an important thing to note about using polar coordinates. Although the unit vectors change over time, the reference frame is inertial. Our perspective is fixed, it does not move or co-rotate with the coordinate system. This an lead to some, at first counterintuitive, results compared to cartesian coordinates. One may assume that if there is no force in the radial direction, then there is no acceleration in the radial direction and $\ddot{r}=0$ . However, as we see above, the radial component of the acceleration vector, which takes into account the change of the unit vectors over time, is $a_{r}=\ddot{r}-r\dot{\theta}^{2}$ . The radial component of acceleration is not the derivative of radial velocity, $\dot{r}$ . This can be confusing because it implies that if $F_{r}=0$ and so $a_{r}=0$ by Newton’s second law, then there can still be motion in the radial direction $\ddot{r}\neq 0$ . We will now look at an example to illustrate this.

Example 4.1.

Consider a bead which is free to move along a frictionless rod. The rod is anchored at one end and is rotating at constant angular velocity $\omega$ . As the rod rotates, the bead will swing out to the end of the rod. Before it reaches the end, what does its motion look like?

Since the total force on the bead is $\boldsymbol{F}=F_{\text{rod}}\hat{\boldsymbol{\theta}}$ , the instantaneous radial acceleration $a_{r}=\ddot{r}-r\dot{\theta}^{2}$ must be zero. This is a differential equation that we can solve for the radial motion.

\frac{\operatorname{\mathrm{d}}\!^{2}r}{\operatorname{\mathrm{d}}\!t^{2}}-% \omega^{2}r=0.

(4.35)

Using an ansatz of the form $r=Ae^{Bt}$ , with two unknown constants $A$ and $B$ , we get

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}}{\operatorname{\mathrm{d}}% \!t^{2}}(Ae^{Bt})-\omega^{2}Ae^{Bt}=0$		(4.36)
	$\displaystyle AB^{2}e^{Bt}-\omega^{2}Ae^{Bt}=0$		(4.37)
	$\displaystyle B^{2}-\omega^{2}=0.$		(4.38)

This gives $B=\pm\omega$ , so we get the general solution

r=A_{+}e^{\omega t}+A_{-}e^{-\omega t}.

(4.39)

When $t=0$ , $r$ is some value $r_{0}$ and $\dot{r}=0$ , so

	$\displaystyle\dot{r}(0)$	$\displaystyle=\left.A_{+}\omega e^{\omega t}-A_{-}\omega e^{-\omega t}\right\|_% {t=0}$		(4.40)
		$\displaystyle=\omega(A_{+}-A_{-})=0,$		(4.41)

therefore $A_{+}=A_{-}$ , so we will relabel them both $A$ . Finally, $r(0)=2A=r_{0}$ , so we have

	$\displaystyle r$	$\displaystyle=\frac{1}{2}r_{0}(e^{\omega t}+e^{-\omega t})$		(4.42)
		$\displaystyle=r_{0}\cosh(\omega t).$		(4.43)