Can a Particle Sliding on a Rotating Rod Decrease in Distance Over Time?

[Saitama]

Homework Statement

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.)

Homework Equations

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!

 

[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?

 

[Saitama]

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?

It is free to move in the radial direction.

What is the other direction? Perpendicular to rod?

 

[BruceW]

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

 

[Saitama]

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

Actually, I still don't realize why second equation should not be zero.

 

[WannabeNewton]

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

 

[ehild]

Actually, I still don't realize why second equation should not be zero.

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

 

[voko]

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

 

[Saitama]

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

Normal force from the rod?

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

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

 

[BruceW]

Normal force from the rod?

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.

 

[ehild]

Normal force from the rod?

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

 

[voko]

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

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?

 

[Saitama]

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

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.

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.

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?

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.

 

[BruceW]

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.

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 $2m\dot{r}\omega$

 

[ehild]

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.

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

 

[Saitama]

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

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?

 

[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

 

[voko]

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.

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.

 

[Saitama]

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.

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. :(

 

[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.

 

[Saitama]

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.

The normal force (N) in the tangential direction balances the force $2m\dot{r}\omega$.

Equating forces in the tangential direction, $N=2m\dot{r}\omega \Rightarrow \dot{r}=N/(2m\omega)$. But the acceleration vector I wrote down gives $\dot{r}=0$. :(

 

[WannabeNewton]

As ehild already mentioned, you're mixing up the corotating frame and the inertial frame. In the inertial frame the only force in the tangential direction is the constraint (normal) force. There is nothing balanced in the tangential direction.

 

[ehild]

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. :(

Think about forces acting on the particle. It interacts with the rod, and with nothing else. The rod does not exert force along its length: "the particle slides freely along the rod". What is the radial component of the net force acting on the particle then?
The force of constraint is perpendicular to the constraint: The normal force is perpendicular to the rod. That force causes the azimuthal acceleration.

I suggest again that you write up the equation F=ma in radial and azimuthal components.

ehild

 

[voko]

Normal force = $mr\omega^2$?

That would be the force felt by the glued particle toward the center, also known as the centripetal force. In the co-rotating frame, that force neutralizes the radial force toward the periphery - the centrifugal force. Now if the particle is no longer glued (but still has to slide along the rod), all it feels is the centrifugal force (the Coriolis force is checked by the reaction of the rod).

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. :(

Do you see what radial force acts on the particle?

Do you see why there is zero net tangential force in the co-rotating frame?

Do you in particular see that tangentially there is a fictitious force, which can be eliminated by switching to an inertial frame, and a real force, which is present in any frame?

What is left of your misconception?

 

[Saitama]

As ehild already mentioned, you're mixing up the corotating frame and the inertial frame. In the inertial frame the only force in the tangential direction is the constraint (normal) force. There is nothing balanced in the tangential direction.

I think I will have to learn more about these frames. I guess I should leave the problem of $\dot{r}=0$ for now and move on to the other parts of the question.

In the first post, it is $\gamma=\omega$. Sorry about the typo.

The next part of the question states that r decreases continually with time. I am not sure how would I approach this but it looks like $Be^{\omega t}$ would increase too much with time so if we set B=0, we can have r which continuously decreases with time but how should I find the initial conditions?

I suggest again that you write up the equation F=ma in radial and azimuthal components.

ehild

For radial direction,
$F_r=ma_r \Rightarrow r\omega^2=\ddot{r}$. The solution of this differential is asked in the question.

For azimuthal direction,
$F_{\theta}=ma_{\theta} \Rightarrow N=2m\dot{r}\omega$. I am sorry if this wrong. Please have a look at my attempt on the other part of the given question.

 

[BruceW]

For radial direction,
$F_r=ma_r \Rightarrow r\omega^2=\ddot{r}$. The solution of this differential is asked in the question.

For azimuthal direction,
$F_{\theta}=ma_{\theta} \Rightarrow N=2m\dot{r}\omega$. I am sorry if this wrong. Please have a look at my attempt on the other part of the given question.

This is correct! :)

 

[Saitama]

Look at the expression of the acceleration.
$\textbf{a}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \hat{ \theta }$
What are its components?

ehild

In the radial direction, $\ddot{r}-r\dot{\theta}^2$.
In the tangential direction, $r\ddot{\theta}+2\dot{r}\dot{\theta}$

 

[ehild]

For radial direction,
$F_r=ma_r \Rightarrow r\omega^2=\ddot{r}$. The solution of this differential is asked in the question.

For azimuthal direction,
$F_{\theta}=ma_{\theta} \Rightarrow N=2m\dot{r}\omega$. I am sorry if this wrong. Please have a look at my attempt on the other part of the given question.

It is OK now.

I think I will have to learn more about these frames. I guess I should leave the problem of $\dot{r}=0$ for now and move on to the other parts of the question.

In the first post, it is $\gamma=\omega$. Sorry about the typo.

It is correct now.

The next part of the question states that r decreases continually with time. I am not sure how would I approach this but it looks like $Be^{\omega t}$ would increase too much with time so if we set B=0, we can have r which continuously decreases with time but how should I find the initial conditions?

The initial conditions mean knowing r and v=dr/dt at t=0. You know the general solution of r(t). You also know that B has to be zero. What are r(0) and v(0) then?

ehild

 

[Saitama]

The initial conditions mean knowing r and v=dr/dt at t=0. You know the general solution of r(t). You also know that B has to be zero. What are r(0) and v(0) then?

ehild

r(0)=A

$r'(t)=-A\omega e^{-\omega t} \Rightarrow r'(0)=-A\omega$

Is this enough? I don't have the final answers.

 

[BruceW]

well, the second part of the question is "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.)" Which you have almost done now. Maybe you could explicitly write out r(0) as a function of r'(0) for the case when r decreases continually with time. Also, you need to give a little bit of explanation for why any other choice will give an eventually increasing r. I think you said "it looks like $Be^{ωt}$ would increase too much with time" This is pretty much the right idea. But maybe make it sound a bit more mathematical. something along the lines of "as t tends to infinity..."

Also, you still need to do the first part of the question.

 

[BruceW]

For radial direction,
$F_r=ma_r \Rightarrow r\omega^2=\ddot{r}$. The solution of this differential is asked in the question.

To do the first part of the question, go from here.

edit: Oh wait, you said you did this bit already. sorry about that.

 

[ehild]

r(0)=A

$r'(t)=-A\omega e^{-\omega t} \Rightarrow r'(0)=-A\omega$

Is this enough? I don't have the final answers.

You can also write that in case r'(0)=-ωr(0) the particle moves inward, that is r(t) decreases all the time . Note, it is interesting. The initial radial speed is the same as the azimuthal one...


ehild

 

[Saitama]

You can also write that in case r'(0)=-ωr(0) the particle moves inward, that is r(t) decreases all the time . Note, it is interesting. The initial radial speed is the same as the azimuthal one...


ehild

Thanks ehild!

 

[kojo90]

Hi everyone, sorry but i am a bit confused as to why radial acceleration is set to 0. Isn't the friction force from the rod pushing the mass inward, accounting for radial acceleration?

 

[haruspex]

Hi everyone, sorry but i am a bit confused as to why radial acceleration is set to 0. Isn't the friction force from the rod pushing the mass inward, accounting for radial acceleration?

What friction?