Polar coordinates and mechanics question.

[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.

Homework Statement

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.

Homework Equations

Now this is the working out that I copied.

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

(I realize 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.

The Attempt at a Solution

\theta = \omegat

\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.

 

[Anadyne]

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.

 

[In_Development]

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.

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