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

[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?

 

[kojo90]

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?

 

[voko]

Yes, "slide freely" means "no friction".