# Homework Help: Polar coords of motion

1. Oct 14, 2006

### Plastik

Question goes something like:
A cylindrical particle is placed in a certain metre long, frictionless tube, where it can slide back and forth. If the particle is not positioned in the centre of the tube, then as the tube is turning at a constant angular velocity of U, determine the acceleration of the particle along the tube.

I understand from my textbook that the equation to use in such case should be
a_r = r "doubledot" (second derivative w.r.t time) - r x U^2
where r = the distance between the particle to the centre of the tube.

But I do not know how to work out r "doubledot".

Any help is much appreciated.

2. Oct 14, 2006

### FunkyDwarf

you need a function that represents the distance r so you can differentiate it twice to get the acceleration (which is r double dot)

3. Oct 14, 2006

### Plastik

But I wasn't provided with any sort of function in relation to r. It's just a particle sliding inside a tube. ><

4. Oct 14, 2006

### FunkyDwarf

ah ok sorry i didnt see the angular velocity bit.

ok

Tangental velocity, v, is equal to angular velocity, w, times radius, r. Ie v/r = w. So if centripetal acceleration, a, is (v^2)/r you should be able to work out a in terms of r and/or w

5. Oct 14, 2006

### Plastik

Oh orite.

So what do I do with the formula

a = r double dot - r*w^2?

can I just ignore it and use a = r*w^2 instead?

6. Oct 16, 2006

### FunkyDwarf

its the same thing :) r double dot is simply the 2nd time derivative of r, which being a spacial value, is acceleration of that value.