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The center of mass stays inside the supports because the bar itself is heavy. To tip over, the center of mass would have to move outside the supports. Here is how you can calculate it: The picture shows a simplified "asymmetrical barbell". The weight of the bar is $W_1$, the weight of the disk added is $W_2$. Now the bar creates a counter-clockwise ...


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Because the barbell is symmetrical, its weight $m_bg$ cannot exert a moment about the point $A$. Standing on its own the barbell would be meta-stable but in combination with the frame and its weight $m_fg$ this is a stable arrangement.


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Ultimately, the third law expresses in Newton's formalism the principle of conservation of momentum. In advanced physics you will understand that the laws of physics of the world can be defined in terms of "Lagrangians" and "action principles". An action principle takes a bunch of trajectories of particles from some starting positions to their ending ...


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At every joint you need the center of mass of the suspended material to be below the support. For mobiles, that is fairly natural.


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The centre of mass exactly below the point of suspension is the equilibrium position. For a small displacement from the equilibrium position you need to have a restoring torque back toward the equilibrium position. Put another way you want the potential energy vs displacement graph around the equilibrium position to exhibit a minimum. Nice description ...


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If a force $\vec{F}=(F_x,F_y)$ is applied at a location $\vec{r}=(x,y)$ then the torque at the origin is $$ \vec{\tau} = \vec{r} \times \vec{F} \\ \tau = x F_y - y F_x$$ All you need to do is sum up the torques at the pivot for the different situations in order to understand how this mechanism will move. If you have two equal and opposite forces $F$ the ...


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If the damping coefficient approaches zero, the differential equation we are looking for needs to approach the 2nd equation you wrote (the one for the physical pendulum). Therefore, we can conclude that the only thing we have to modify in the last equation to get the equation for a damped physical pendulum is to change g/L into mgL/(I_CM+mL^2). If we want ...


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A physical pendulum as described above behaves identically to a simple pendulum with a length $L^\prime=\frac{I_{\rm{CM}}+mL^2}{mL}$. So my inclination would be that you just replace both L's in your equation with the $L^\prime$ I just defined. That is: $\frac{\partial^2\theta}{\partial t^2}+\left(\frac{\xi m ...



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