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Introductory Physics Homework Help
Small oscillation frequency of rod and disk pendulum
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[QUOTE="Jenny Physics, post: 6075342, member: 604610"] [h2]Homework Statement [/h2] Consider a rod of length ##L## and mass ##M## attached on one end to the ceiling and on the other end to the edge of a disk of radius ##r## and mass ##m##. This system is slightly moved away from the vertical and let go. Let ##\theta## be the angle the pendulum makes with the vertical. What is the frequency ##\omega## of small oscillations of this pendulum? [ATTACH=full]232515[/ATTACH] [h2]Homework Equations[/h2] Newton's rotation law gives $$\vec{\tau}=I\ddot{\theta}$$ where ##I## is the moment of inertia of the rod and disk relative to the pivot in the ceiling and ##\vec{\tau}## is the torque due to the gravity force. [h2]The Attempt at a Solution[/h2] [/B] The torque due to gravity around the pivot ##P## divides into the torque of gravity at the center of mass of the rod (located at ##L/2##) plus the torque of the gravity force at the center of mass of the disk (located at ##L+r##). This is $$\vec{\tau}=\vec{r}_{P,M}\times M\vec{g}+\vec{R}_{P,cm}\times m\vec{g}$$ This ends up giving $$\vec{\tau}=-\left[M\frac{L}{2} +m(L+r)\right]g\sin\theta\hat{z}$$ So the equation of motion $$\vec{\tau}=I\ddot{\theta}$$ becomes $$-\left[M\frac{L}{2} +m(L+r)\right]g\sin\theta=I\ddot{\theta}$$ If we make this an harmonic type equation where ##\ddot{\theta}=-\omega^{2}\theta,\sin\theta\approx \theta ## it becomes ##-\left[M\frac{L}{2} +m(L+r)\right]g\theta=-I\omega^{2}\theta## and so $$\omega=\sqrt{\frac{\left[M\frac{L}{2} +m(L+r)\right]g}{I}}$$ This is not the solution for the frequency of oscillation, which means I am missing something perhaps in the center of mass calculation. [/QUOTE]
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Introductory Physics Homework Help
Small oscillation frequency of rod and disk pendulum
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