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

[srecko97]

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

 

[TSny]

Keep terms of first order in φ. cosφ ≈ 1 - φ²/2 ≈ 1 to first order. What about sinφ?

 

[srecko97]

sinφ=φ-(φ^3)/6

 

[TSny]

sinφ=φ-(φ^3)/6

So, to first order in φ?

 

[srecko97]

oh it is φ

 

[TSny]

oh it is φ

That's right.

 

[srecko97]

Well, I got that M_right - M_left = mg(rφ-bφ) ...M=torque

 

[srecko97]

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...

 

[TSny]

Your answer looks correct to me. (b ≥ r is interesting to think about.)

Glad I could help.

 

[srecko97]

ω² is then negative, so there is no oscillation in real?

 

[TSny]

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.