Spring-mass system in circular motion has a maximum angular speed?

  • #1

Homework Statement:

It is a mass doing circular motion on a smooth horizontal table.
For a given maximum tension ##F_(max)## , I can solve for a maximum angular speed ##\omega_c## of circular motion. But I don't understand how Under some conditions the system can only achieve a maximum angular frequency ##ω_i < ω_c##.
I don't understand part (d)

Relevant Equations:

Hooke's law $$F=k(R-R_0)$$
Tension provides as centripetal force $$k(R-R_0)=m\omega^2R$$
pic1.png

and this is my solution
WechatIMG17573.png

for question (d), it may seems that $$R=(k)/(k-m\omega^2)R_0$$ so that $$\omega ≠ \omega_i =√(k/m)$$
but $$\omega_c <\sqrt{k/m}$$ is always true, ##\omega_i## corresponds to the limit case when ##F_max## is infinitely large
Besides, I don't know other Physics prevents ##\omega## reaches ##\omega_c##
 

Answers and Replies

  • #2
etotheipi
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I think you're going along the right lines. Look at your expression for $$R = \frac{k}{k-m\omega^2}R_0$$Are there any values for ##k## and ##m## for which that expression is problematic? If ##\omega## exceeds ##\omega_i##, does the circular motion still hold, or do we get something else?
 
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  • #3
I think you're going along the right lines. Look at your expression for $$R = \frac{k}{k-m\omega^2}R_0$$Are there any values for ##k## and ##m## for which that expression is problematic? If ##\omega## exceeds ##\omega_i##, does the circular motion still hold, or do we get something else?
Thank you my first replay here in PF:)
Yes, but what I can see is only that the denominator cannot be 0. i.e. $$k-m\omega^2≠0$$ so you are suggestion the limit is ##\omega_i^2=\frac{k}{m}##?
But there is always $$\omega_c<\omega_i=\sqrt{\frac{k}{m}}$$ and the question requires us to find a limit ##\omega_i##<##\omega_c##
 
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  • #4
etotheipi
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Thank you my first replay here in PF:)
Yeah, welcome :smile:

Yes, but what I can see is only that the denominator cannot be 0. i.e. $$k-m\omega^2≠0$$ so you are suggestion the limit is ##\omega_i=\frac{k}{m}##?
But there is always $$\omega_c<\omega_i=\frac{k}{m}$$ and the question requires us to find a limit ##\omega_i##<##\omega_c##
Right. The ##\omega^2 = \frac{k}{m}## is the resonant case, and corresponds to where ##R \rightarrow \infty## (not to mention, the spring will break!). So you won't find any solutions for circular motion ##\omega^2 > \frac{k}{m}##. That doesn't mean you won't find more exotic solutions, though!

At ##\omega^2 < \frac{k}{m}##, you will still be able to find circular motion solutions.

I don't know what the question means with ##\omega_i < \omega_c##, since clearly you can show this is not true. Let us wait for someone else to add their opinion!
 
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  • #5
haruspex
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A stability issue, perhaps?
 
  • #6
etotheipi
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Also. I think there is an inconsistency in the naming of variables here. You, and I initially, both took ##R## to be the current radial coordinate, which leads to the equation$$k(R-R_0) = mR\omega^2 \implies R = \frac{k}{k - m\omega^2}R_0$$and then if we are to set ##\omega_i^2 := \frac{k}{m}##, we would obtain$$\omega^2 = \frac{R-R_0}{R}\omega_i^2$$However the original question seems to say that ##R## is actually the extension, in which case the equation of motion is actually$$kR = m(R+R_0)\omega^2 \implies R = \frac{mR_0 \omega^2}{k-m\omega^2}$$Now if we set ##\omega_i^2 := \frac{k}{m}## we get$$\omega^2 = \frac{R}{R+R_0} \omega_i^2$$A small but nonetheless important detail; that will probably change your answers to b) and c) a little bit?

You can see that ##\ddot{\theta} = 0## since there are no external torques, and from NII$$m\ddot{r} - mr\dot{\theta}^2 = -k(r-R_0)$$Now ##mr\dot{\theta}^2 = \frac{L^2}{mr^3}##, so$$m\ddot{r} = -k(r-R_0) + \frac{L^2}{mr^3}$$i.e. the situation is equivalent to a particle moving in an effective potential$$U_{\text{eff}}(r) = \frac{L^2}{2mr^2} + \frac{1}{2}k(r-R_0)^2$$so there is definitely more exotic motion available!
 
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  • #7
Also. I think there is an inconsistency in the naming of variables here. You, and I initially, both took ##R## to be the current radial coordinate, which leads to the equation…
Yeah, you are right, thanks. I didn't notice R is actually extension here, so expression for R has to change. Besides, the graph given in question is also misguiding.
But this won't affect my answer b) and c) since I didn't use formula of extension R when I derive answer in b) and R never gets to appear in answer there.

Still, I am puzzled by part d) where ##\omega_i## < ##\omega_c##
 
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  • #8
etotheipi
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Still, I am puzzled by part d) where ##\omega_i## < ##\omega_c##
If it is any consolation, I too am puzzled. ##\omega_i## and ##\omega_c## are constants (for fixed ##k##, ##m##, ##R_0##, as is to be assumed), so obviously one is going to be greater than the other always.

I suspect it actually means to say that ##\omega < \omega_i##... that would make the most sense given the context, but it's a little different to what's actually written ?:)

I say mistake - would you agree @haruspex?
 
  • #9
haruspex
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If it is any consolation, I too am puzzled. ##\omega_i## and ##\omega_c## are constants (for fixed ##k##, ##m##, ##R_0##, as is to be assumed), so obviously one is going to be greater than the other always.

I suspect it actually means to say that ##\omega < \omega_i##... that would make the most sense given the context, but it's a little different to what's actually written ?:)

I say mistake - would you agree @haruspex?
I still think it smells like a stability issue. At some critical angular rate, small perturbations might lead to wild swings, potentially breaking the spring.
I am not accustomed to solving stability questions though, so I do not straight away see how to analyse it.
I think I managed to an equation relating ##\dot r^2, L, r##, where L is the (constant) angular momentum, but has both an ##r^2## term and an ##r^{-2}## term.
 
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  • #10
Could it possibly be the condition that it is not doing steady circular motion (if we increase the angular frequency very slowly to its maximum), but instead, we give the ball a tangential impulse at first. So the ball will start doing circular motion with angular frequency ##\omega_i## but spring will extend, could it potentially break the spring with such a critical initial ##\omega_i## which turns out to be smaller than ##\omega_c##?
by conservation of Angular momentum $$R_0 \times \omega_i R_0 = r \times \omega r$$ by conservation of energy $$\frac{1}{2}m\omega_i^2 R_0^2=\frac{1}{2}m(\frac{\omega_iR_0^2}{r})^2+\frac{1}{2}k(r-R_0)^2$$ if we set ##r=R_{max}=\frac{F_{max}}{k}+R_0## we can get ##\omega_i##

I just don't know would this method work and would ##\omega_i < \omega_c## be possible
 

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