OK, the answer for this problem seems a bit high to me, so I'm going to ask if it all seems alright.(adsbygoogle = window.adsbygoogle || []).push({});

You have a 1m rod of no mass, fixed so that it may rotate about it's center. At the top of the rod is a mass m1 (0.5kg), and the bottom is a mass m2 (1.0kg). Find the angular frequency assuming a small amplitude of oscillations.

I just stared at it for about an hour, trying to find a way to get it into the form

[tex]\ddot{\psi} + \omega_0 ^2 \psi = 0[/tex]

Eventually i went at it with torques, using the [tex]\tau = r x F[/tex] forumla.

So...

[tex] \tau = rm_1g \sin (180- \theta) - rm_2g \sin (\theta) [/tex]

and using small angle approximation, both sins go to theta.

[tex] \tau = rg (m_2 - m_1) \theta [/tex]

[tex] \tau = I \ddot{\theta} = -rg (m_2 - m_1) \theta [/tex]

[tex] I = \sum m r^2 = (m_1 + m_2) r^2 [/tex]

[tex] \ddot{\theta} = \frac{-rg (m_2 - m_1)}{r^2 (m_1+m_2)}[/tex]

[tex] \ddot{\theta} + \frac{rg (m_2 - m_1)}{r^2 (m_1+m_2)} = 0 [/tex]

Which is in the right form. Meaning that the frequency is:

[tex] \omega_0 = \sqrt{\frac{g(m_2-m_1)}{r(m_1+m_2)}} [/tex]

Now, that may be ALL wrong, but it's what first came to mind. I end up with 2.6Hz for an answer which seems a bit off...

Thanks for any redirections/confirmations/help you can give!

(if the tex is messy i'm working on it still!)

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# Not so simple harmonic motion

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