1. Mar 15, 2013

### cacofolius

1. The problem statement, all variables and given/known data

A particle of mass m is free to slide on a thin rod. The rod rotates in a plane about one end at a constant angular velocity w. Show that the motion is given by r=Ae^(-γt)+Be^(γt), where γ is a constant which you must find and A and B are arbitrary constants. Neglect gravity.
Show that for a particular choice of initial conditions (that is, r(t=0), and v(t=0) ), it is possible to obtain a solution such that r decreases continually in time, but that for any other choice r will eventually increase. (Exclude cases where the bead hits the origin).

2. Relevant equations

a=(r''-rw^2)êr + (rθ''+2r'we) êθ

3. The attempt at a solution

Ok, for the first part I took twice the derivative of r, and plugged in the radial part of the acceleration (being θ'' and the coriolis force equal to zero, in the stationary frame of reference).
So I only got r''=rw^2, from which I deduced that γ must be w.

From the initial conditions, where t=0, I get

r(0)=A+B=r0
r'(0)=w(B-A)=0 (since I assume it starts with no initial speed), so A=B

From these I get that A=B=r0/2, so the complete solution for the position is

r=(r0/2) [e^(-wt)+e^(wt)]

Now, to show that there is a solution where r decreases in time, I figured there should be an initial speed towards the origin, so

r'(0)=w(B-A) does not equals zero, but a certain initial velocity and so

r'(0)=w(B-A)=Vo means

B=(Vo/w)+A

Therefore, the case where r decreases in time should satisfy

r=Ae^(-wt)+[(Vo/w)+A]e^(wt)

Is this correct, or is there another way of showing this ?

2. Mar 15, 2013

### TSny

Looks good so far. You have not yet found the explicit solution where r decreases continuously in time.

3. Mar 18, 2013

### cacofolius

Is it possible then to choose A such as A=-Vo/w? That way I'll have just r=Ae^(-wt) which decreases with time.

4. Mar 18, 2013

Yes. Good.