Finding frequency of a specific mechanical oscillator -- horizontal rod on pivot

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Homework Help Overview

The discussion revolves around calculating the oscillation period of a mechanical oscillator, specifically a thin rod placed on a cylinder. The original poster expresses confusion over their calculated period, which is significantly longer than expected for a clock mechanism.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the moment of inertia and whether to treat different sections of the rod separately. There are inquiries about the effects of the rod's width on the moment arms and the overall calculations.

Discussion Status

Several participants have confirmed the correctness of the original poster's work while also suggesting alternative approaches. There is ongoing exploration of the implications of the rod's width and the moment arms involved in the calculations.

Contextual Notes

Participants note the lack of specific data regarding the width of the rod and its impact on calculations. The original poster expresses uncertainty about the relevance of their findings to clock mechanisms, indicating a potential misunderstanding of the problem's context.

  • #31
srecko97 said:
On the other side it gets shorter for b*sinφ
Yes.
 
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  • #32
But when talking about small φ that has no sense
 
  • #33
srecko97 said:
But when talking about small φ that has no sense
Why is that?
 
  • #34
sin φ ≈ 0 for small φ
 
  • #35
srecko97 said:
sin φ ≈ 0 for small φ
That's going too far in the approximation.
 
  • #36
I did the same approximation in the first part of my homework (calculating the oscillating time), when I say cos φ≈1 for small φ
 
  • #37
Keep terms of first order in φ. cosφ ≈ 1 - φ2/2 ≈ 1 to first order. What about sinφ?
 
  • #38
sinφ=φ-(φ^3)/6
 
  • #39
srecko97 said:
sinφ=φ-(φ^3)/6
So, to first order in φ?
 
  • #40
oh it is φ
 
  • #41
srecko97 said:
oh it is φ
That's right.
 
  • #42
Well, I got that M_right - M_left = mg(rφ-bφ) ...M=torque
 
  • #43
image.jpg


The frequency gets smaller, as numerator gets smaller and denominator gets bigger. I hope it is correct! I would like to say you a big THANK YOU TSny! You helped me a lot... I learned a lot from solving this task with your help. I love this forum...
 
  • #44
Your answer looks correct to me. (b ≥ r is interesting to think about.)

Glad I could help.
 
  • #45
ω2 is then negative, so there is no oscillation in real?
 
  • #46
srecko97 said:
ω2 is then negative, so there is no oscillation in real?
Yes, there is no oscillation. Looking back at your post#43, I think you need to add an overall negative sign to the first equation so that it's

##J \ddot{\phi} = - mg(r-b)\phi##.

If ##b > r## then you can see that ##\ddot{\phi} >0## if ##\phi## is given an initial small, positive value. So, the rod rotates away from the equilibrium position when it is released and there is no oscillation. Of course, the differential equation for ##\phi## breaks down as ##\phi## continues to increase since the differential equation was derived under the assumption that ##\phi## is always small.

Another way to investigate the general behavior of the system for various values of b is to derive the potential energy of the system as a function of ##\phi## without making a small angle approximation. Then plots of the potential energy function for various values of b will reveal a lot about the behavior. But that's probably for a rainy day.
 
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