Angular acceleration - radial & tangential

Question

Since ever I knew that radial (angular acceleration) is equal to $ W^2 * R = V^2 / R $ and that the tangential depends in the situation (School physics & calculus). Recently I encountered the following:

I understand in the first line the term $-r*w^2$, and I familiar with the expression in the second line , $r*a$. What about the others? I would like to know how they are derived. I want also to understand the meaning of the first and second derivative of $r$.

Answer 1

In polar coordinates you have $(x,y) = (r \cos \theta, r \sin \theta)$

Taking total derivatives of the above one finds that:

• Positions $$\begin{aligned} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \begin{pmatrix} r \\ 0 \end{pmatrix} \end{aligned} $$
• Velocities $$\begin{aligned} \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} &= \begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \begin{pmatrix} \dot{r} \\ r \dot{\theta} \end{pmatrix} \end{aligned} $$
• Accelerations $$\begin{aligned} \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} &= \begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \begin{pmatrix} \ddot{r}-r \dot{\theta}^2 \\ r \ddot{\theta} + 2 \dot{r}\dot{\theta} \end{pmatrix} \end{aligned} $$

The $\ddot{r}-r \dot{\theta}^2$ is explained as the net radial acceleration to keep the object moving in a curved line. The $r \ddot{\theta}$ part is the tangential acceleration because the angular velocity changes, and the $2\dot{r}\dot{\theta}$ part is the tangential acceleration because the position $r$ changes affecting the angular momentum.

Answer 2

If I understand your problem right, you want to see why the terms are radial/tangential:

Radial

$-r \dot{\theta}^2$ is indeed the term you know from a uniform circular motion. As in that case (UCM) only the direction of the velocity changes, the acceleration is perpendicular to the velocity, thus in the radial direction.

$\ddot{r}$ if the radius is changing and the rate of this change is changing as well, this is it, so it's just the change in radial velocity. It's quite logical that this part of the acceleration is in the radial direction.

Tangential

$r\ddot{\theta}$ a change in angular velocity while keeping the radius constant. As in this case (constant radius) the movement is circular, a change in angular velocity resuls in a change of the magnitude of the velocity, so the acceleration must be tangential.

$2\dot{r}\dot{\theta}$ consider something in a uniform circular motion ($\dot{\theta}$, but no $\dot{r}$). If suddenly, the radius is changed, the magnitude of the velocity is changed ($v=r\dot{\theta}$, if $r$ changes, $v$ changes), so there must be a tangential component to the acceleration.

Answer 3

Your familiar equation of $a = r\alpha + r\omega^2$ are for circular motion with a constant radius, i.e. swinging a mass around on a string. As you change the angular velocity ($\alpha$ is nonzero), the velocity of the mass in the direction of its motion (tangent to the circle of radius $r$) will change as well. In addition, we learn in any physics class that an object moving in circular motion is accelerating towards the center (radially) of the circle at $r\omega^2$.

Now, replace that string with either a rubber band or a slider, so the path of the mass is no longer confined to the circle of radius $r$. If the length of the radius is changing at a non-constant rate, then the mass is accelerating along the radius (if you attached a camera or a motion sensor to the base of the rubber band/rope/slider, you would see the mass accelerating towards you or away from you).

The last term comes from the coriolis effect, which appears in a rotating system with a changing radius. Because the mass' tangential speed is proportional to both its angular velocity and its radius, a change in radius will cause a change in its tangential velocity (assuming a constant angular velocity).