Question

I'm having some trouble seeing how my professor derived the acceleration around a circle using polar coordinates. I'll attach my notes below. Thanks! :)

Answer 1

Actually, this time I'll upload it right-side up.

Answer 2

It's the part outlined in orange that's giving me grief. First, I think I understand how velocity in a circle could be\[\frac{ dv }{ dt } (r \space unit)+\frac{ d \theta }{ dt } (r \space \theta unit)\] What is confusing me is how he got to the acceleration with \[2 \frac{dv}{dt}(r\space unit)\]

Answer 3

Hi, without looking at the details, you get the acceleration by differentiating the velocity with respect to time, as I'm sure you know.
So that first line you have is the velocity in polar coordinates, fine.
Now, when you differentiate each of those terms, remember that they are themselves products, so you have to use the product rule.
That will give you 4 terms in the expression for the acceleration, and I think two of them happen to be the same so they add up to give you the term you're worrying about.
If I have time I'll check the details.

Answer 4

Sorry, slight mistake above, that second term is a product of 3 factors, so you will generate 5 terms in total for the acceleration when you differentiate v.

Answer 5

Ok, so here's what I have so far:

Answer 6

yes that looks correct so far
now you just need to differentiate those unit vectors and you should have it

Answer 7

I'm not sure I follow you on that last part. Aren't they differentiated already in this form?

Answer 8

the unit vectors r and theta are not fixed in the polar coordinate system

Answer 9

Eeeks. :/

Answer 10

actually, i'm not sure where you've got this 2dv/dt r unit term from ? it doesn't appear in those first notes you posted

Answer 11

It's on the third line of the very first part, outlined in orange.

Answer 12

if you differentiate r unit with respect to time, you get dtheta/dt times theta unit
and if you differentiate theta unit, you get -dtheta/dt times r unit

Answer 13

Aaah...ok, give me a second to try that.

Answer 14

Just to make sure I have the intuition, then, the change in the r-unit vector over time is equal to the change of the angle theta with respect to time, but in the direction of the theta-unit. So the theta-unit becomes the direction for the change in angle over time?

Answer 15

Well the first part is correct, I'm not sure what you mean by direction of a change in angle though.

Answer 16

do you have the expression for acceleration okay now ?

Answer 17

Me either. :) I'm almost to acceleration. I'll post in a sec. I didn't want to just plug-and-chug.

Answer 18

The way to see the differentiation of the vectors is to draw a diagram and see how a small change in r affects the unit vectors, then separately consider how a small change in theta affects the unit vectors, that should make it clear

Answer 19

you'll see that only the changing angle has an effect on the unit vectors

Answer 20

So the last line is where I am now.

Answer 21

you've lost me there (it's late here, my brain's struggling) it looks like you might have made a mistake somewhere

Answer 22

yes, i've looked again
in your last line, the third term is missing a factor of r dot
and the very last term shouldn't be there at all
then you've got it

Answer 23

when you put in the missing factor of r dot in the third term, you'll see that the third and fourth terms are actually the same , that's where you get the 2 from : )

Answer 24

Had to help the kid for a sec, sorry. Working through it now.

Answer 25

Ah! nice catch. I should probably have done this part in Mathematica. Unfortunately, I had been using Sage, and haven't had time to learn it yet. Ok, this should do it...

Answer 26

Sage ? Mathematica ?
pen and paper for me : )

Answer 27

Usually, yes, me too. It's an honors course, so I think he's trying to cram in as much as possible. :)

Answer 28

Good luck with it all

Answer 29

Thanks. One thing--in my last line, won't they cancel out? I thought it was:\[\frac{d\theta(unit)}{dt}=-\frac{d\theta}{dt}r(unit)\]

Answer 30

Oh, duh. I missed the second part of what you said. Reworking...I think I've got it now, thanks!

Answer 31

ok, good