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I think that @Dhiraj_Barnwal is questioning as to whether or not the problem can be solved and so it is a valid query. If you consider any of the answers to be true and conserve angular momentum then energy conservation is violated. If angular momentum is not conserved then there must be a external torque acting on the system of three cylinders and that ...


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It says "no slip". That means more than enough friction, so you can treat the string as if were a chain draped over a sprocket. They're trying to simplify the problem for you.


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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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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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The torque is done with a cross product and thus is going to be:$$\tau = |\vec F| |\vec r| \sin\theta$$where $\theta$ is the angle between the position vector $\vec r$ (which points from the pivot to the place where the force is applied) and $\vec F$ (which points however the force points). You can also express this as $x F_y - y F_x$ and since your forces ...


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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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$F_x$ and $F_y$ are components of a force. If the line of action of that force goes through the bob then the torque that the forces exerts about the bob is zero. In the text those two component forces are external forces applied to the bob via a connecting wire which presumably can only transmit tensile and compressive forces ie forces along the wire.


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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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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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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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Symmetry, no direction is in any way special so the particles cannot decide which direction the torque can it be, spontaneous symmetry breaking is impossible because this torque apparently is large before anything moves. Also, this will violate conservation of angular momentum because the angular momentum around the centre of mass will change (if one mass ...


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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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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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