# Polar coordinates and mechanics question.

1. Oct 21, 2007

### In_Development

Alright, the problem here is that I seem unable to grasp an example given in class. I am not sure if this is due to not copying it down correctly, or if there's something I am just missing. Either way, I know I am not the only one who has had a bit of trouble with this. I'm hoping that someone with a greater knowledge of physics can spot what is going on.

1. The problem statement, all variables and given/known data
Bead threaded on a smooth rod (i.e. no friction) which rotates with constant angular velocity $$\omega$$ about one end. Investigate the motion.

Where r is a radial vector,

$$\dot{r}$$=0, $$\dot{\theta}$$=$$\omega$$, and r=r$$_{0}$$ at t=0

The dot above the letters and symbols, as far as I am aware, just denotes the first derivative (in case this isn't a standard notation of some sort). Also, the subscript doesn't seem to be working in the preview, so in case it does not turn out right, that is meant to be a subcript zero, not r^0. It's just denoting a particular distance up the rod that the bead is.

2. Relevant equations
Now this is the working out that I copied.

Newtons second law:
$$\ddot{r}$$ -r$$\dot{\theta}$$^2 = 0.

(I realise Newton's second law should include a multiplication by mass, but I'm just writing the information I have). r dot dot is the second derivative I believe.

3. The attempt at a solution

$$\theta$$ = $$\omega$$t

$$\dot{\theta}$$ = $$\omega$$ ; $$\ddot{r}$$ = $$\omega$$*r$$^{2}$$

Solutions:
e^wt and e^-wt

Oh, the omega is not meant to be superscript throughout the question by the way, I don't know why it is; and the w in the 'solution' represents an omega. So it is e to the power of omega*time.

All I need to know is what on Earth the solutions are for and how did they materialise?

EDIT: I really made a mess of this whole latex thing.

Last edited: Oct 21, 2007
2. Oct 21, 2007

Let ω=omega θ=theta, r' = dr/dt, r'' = d^2r/dt^2 (second derivative) Let R be the vector , and r be the length.
I'm not sure how you got to Newton's second law but here's how my professor proved it:
R can be broken down into x and y components as such:
R = rcosθ*i + rsinθ*j (i,j, are unit vectors along x and y)

Let's take the derivative twice (Keep in mind that this is with respect to t)
R' = -rsinθ*θ'*i + rcosθ*θ*'j
R'' = -rcosθ*θ'^2*i - rsinθ*θ'^2*j

Hey! That looks like our original R almost:
R'' = -Rθ''
R'' + Rθ'^2 = 0 <---matches your notes except for the sign..
Keep in mind that θ' = ω so:
R'' + R(ω^2) = 0

I haven't taken differential equations and I'm guessing you haven't either. But in general if you want to solve for R but you know that its second derivative is itself times a constant, then there are only a few functions that do that: sine, cosine, and e^x. You "guess" that the solution is e^ωt. If you take the derivative twice, indeed, it is itself times a constant (ω^2).

The solutions are for R. Sorry I don't know why I had issues with the signs, but they should be right otherwise.

3. Oct 22, 2007

### In_Development

Ah thanks for the help, because my professor gave no explanation whatsoever on how he got from one step to the other.