# Particle sliding on a thin rod

1. Aug 12, 2013

### Pranav-Arora

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 constant angular velocity $\omega$. Show that the motion is given by $r=Ae^{-\gamma t}+Be^{+\gamma t}$, where $\gamma$ 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

3. The attempt at a solution
I guess I have to use polar coordinates. In polar coordinates,
$$\textbf{a}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \hat{ \theta }$$
Here, $a=0$ and $\ddot{\theta}=0$. Hence
$$0=(\ddot{r}-r\omega^2)\hat{r}+2\dot{r}\omega\hat{\theta}$$
This gives,
$\ddot{r}-r\omega^2=0$ and $2\dot{r}\omega=0$.
Solving the first equation gives a solution of the form presented by the question and I get $\gamma=\sqrt{\omega}$ but the second equation gives $\dot{r}=0$. This doesn't look right.

Any help is appreciated. Thanks!

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2. Aug 12, 2013

### BruceW

It says 'the particle is free to slide on the rod'. What direction is it free to move in? And what does that imply about the other direction?

3. Aug 12, 2013

### Pranav-Arora

It is free to move in the radial direction.

What is the other direction? Perpendicular to rod?

4. Aug 12, 2013

### BruceW

yeah. I'm sure you'll be like 'oh of course' when you realize why equation 2 is not equal to zero.

5. Aug 12, 2013

### Pranav-Arora

Actually, I still don't realise why second equation should not be zero. :uhh:

6. Aug 12, 2013

### WannabeNewton

If it were zero, then what is keeping the particle constrained to move along the rod?

7. Aug 13, 2013

### ehild

If the total acceleration is zero the particle moves along a straight line with constant velocity. (Newton's First Law. )
Your formula for the acceleration refers to an inertial frame of reference, using polar coordinates.

ehild

8. Aug 13, 2013

### voko

This problem is easiest in the co-rotating frame. What forces act on the particle? What forces does the rod neutralize?

9. Aug 13, 2013

### Pranav-Arora

Normal force from the rod?

Sorry, I don't know about the co-rotating frame.

10. Aug 13, 2013

### BruceW

hehe, I think you have a mind-block on this problem. You are saying that the normal force from the rod keeps the particle constrained to move along the rod and you are saying there is zero force acting on the particle. These two things are inconsistent.

edit: except it is not inconsistent in the special case when omega is zero.

11. Aug 13, 2013

### ehild

The particle slides freely along the rod, but it must stay on the rod. Assume it is a bead with a hole in the middle and the rod goes across the hole. The road exerts a normal force on the bead.

ehild

12. Aug 13, 2013

### voko

This is the frame in which the rod is stationary. It will have fictitious (inertial) forces. The frame rotates uniformly, so these forces are well known.

But even if you do not know them, these forces will have two components: along the rod and perpendicular to it. Do you care about the perpendicular component? Why? What about the parallel component?

13. Aug 13, 2013

### Pranav-Arora

I meant that the force $2m\dot{r}\omega$ (coriolis force?) is balanced by the normal force exerted by the rod on the particle or bead.

The particle is constrained to move along the length of rod. The normal force balances the perpendicular component. The parallel component is responsible for the motion of particle along the rod.

14. Aug 13, 2013

### BruceW

as you said, $\ddot{\theta}$ is zero, therefore $2m\dot{r}\omega$ is the net force on the particle in the angular direction. This is not a fictional force, or anything like that. When you add up all the real forces in the angular direction, they must equal $2m\dot{r}\omega$

15. Aug 13, 2013

### ehild

You mix the two frames -the inertial and the rotating ones. There is no Coriolis force in the inertial frame of reference, but there is normal force from the rod.

In the rotating frame of reference, you have fictitious forces: the centrifugal force and the Coriolis force. And you also have the real normal force from the rod. But the bead moves only radially in that system, so dθ/dt=0.

ehild

16. Aug 13, 2013

### Pranav-Arora

Okay, let's not talk about the other reference frames and solve it in the inertial frame. I haven't yet studied the Coriolis force so I think it would confuse me more if I try to work on it using the rotating frame.

But how do I solve this in inertial frame now? Hmm...I think my acceleration vector is wrong. I did not take into account that the normal force also acts in the $\hat{θ}$ direction. Correct?

17. Aug 13, 2013

### ehild

You wrote the acceleration vector correctly, but did not wrote the equation F=ma. If you write it, you need to include the normal force to the left-hand side.

ehild

18. Aug 13, 2013

### voko

Correct, so you only care about the parallel (radial) force. Still pretending you do not know what it is, assume that the particle is glued to the rod at some radius. Then it is stationary in the rotating frame, consequently the radial force must be neutralized by the normal force toward the center. I am sure you do know what this normal force toward the center is - just consider that particle in the inertial frame now.

19. Aug 13, 2013

### Pranav-Arora

Normal force = $mr\omega^2$?

I don't really understand where this discussion is heading. I asked about the wrong result produced but I don't think I am anywhere close to clearing my misconception. :(

20. Aug 13, 2013

### WannabeNewton

Go back to the inertial frame. The particle can slide freely along the length of the rod. This means it is moving freely in the radial direction but clearly there is a constraint force acting on the particle in the tangential direction that is keeping it constrained to move along the length of the rod.