Particle sliding on a thin rod

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  • #26
BruceW
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For radial direction,
[itex]F_r=ma_r \Rightarrow r\omega^2=\ddot{r}[/itex]. The solution of this differential is asked in the question.

For azimuthal direction,
[itex]F_{\theta}=ma_{\theta} \Rightarrow N=2m\dot{r}\omega[/itex]. I am sorry if this wrong. Please have a look at my attempt on the other part of the given question.
This is correct! :)
 
  • #27
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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}##
 
  • #28
ehild
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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? :confused:
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
 
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  • #29
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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.
 
  • #30
BruceW
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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 [itex]Be^{ωt}[/itex] 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.
 
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  • #31
BruceW
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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.
 
  • #32
ehild
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: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
 
  • #33
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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! :smile:
 
  • #34
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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?
 
  • #35
haruspex
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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?
 
  • #36
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well, just like the rod produces a perpendicular contact force (N), i would assume, as the ball wants to move radially out, there would be some friction involved. Or does the fact that the problem says, "slide along the rod freely" tell us to ignore this?
 
  • #37
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Yes, "slide freely" means "no friction".
 

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