# Particle sliding on a thin rod

BruceW
Homework Helper
$F_r=ma_r \Rightarrow r\omega^2=\ddot{r}$. 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! :)

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 tangential direction, ##r\ddot{\theta}+2\dot{r}\dot{\theta}##

ehild
Homework Helper
##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

• 1 person
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
:tongue:
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
Homework Helper
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.

• 1 person
BruceW
Homework Helper
##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
Homework Helper
:tongue:
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

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! 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