# Homework Help: Tangential and radial acceleration

1. Apr 17, 2012

### BitterX

Hey

I have an accelerated circular motion problem.

I have only the position equation, from which I derived the velocity and acceleration.

how can I tell what is the tangential acceleration and what is the radial acceleration?

If you could point me towards a source to read about the subject, it would be great.

2. Apr 17, 2012

### dikmikkel

The tangential is the usual position, velocity or acceleration. The radial is called the angular postition,velocity or acceleration
So they are connected by
$v = \omega * r , a = r*\alpha$

3. Apr 18, 2012

### BitterX

Let's see if I understand, here's an example:

I have:
$a= -40\pi [ \sin (2\pi t^2 - \frac{\pi}{3}) + 4\pi t^2 \cos (2\pi t^2 - \frac{\pi}{3})] \hat{i} + 40\pi [\cos (2\pi t^2 - \frac{\pi}{3})- t^2 \sin (2\pi t^2 - \frac{\pi}{3})] \hat{j}$

(I derived it from $r= 10 \cos (2\pi t^2 - \frac{\pi}{3}) \hat{i} + 10\sin (2\pi t^2 - \frac{\pi}{3}) \hat{j}$ )

I was asked to determine what is the acceleration without direction at $t=\sqrt{2}$

and also the tangential and radial acceleration.

so I just plugged it in, and got about 483.5 m/s^2

so by what you say, $\sqrt{a^2 + (\frac{a}{R})^2} = 483.5$ ?

edit: I looked in wikipedia under Non-uniform circular motion, I understand now - the value I got was for the tangential acceleration, the radial is simply $\frac{v^2}{R}$ I don't know why I thought it's only true for Uniform circular motion

Last edited: Apr 18, 2012