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3

At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...


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Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved.


2

The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ...


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In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ...


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If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.


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The error lies in the theoretical confusion between forces/torques and energy. The kinetic energy is linked to the motion generated by forces and torques which are the causes of the motion itself. Understanding the energy value in the two situations results impossible without knowing the time course of the applied force, being the energy, and so the work, ...


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You are almost correct. The linear acceleration of the end of the rod is αL perpendicular to the rod. The component of this acceleration in the direction of the string is αLsinA where A is the angle which the string makes with the rod (not the angle which the rod makes with the horizontal). So at the instant shown in the diagram the component along the ...


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First, there are a few different methods developed to solve this kind of problem, but it's highly dependent on your background knowledge (calculus), and experience with these kinds of problems. This approach might be over-detailed for some mechanics problems, but this approach is fairly general, so should usually work. Identify any clues in the problem as ...


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shouldn't we use the parallel axis theorem ... to compute the moment of inertia? You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis.


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You have made some errors in you calculation of distances. \begin{aligned} r_1^2 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1872}{121}\quad \text{(upper left particle)}\\ r_2^2 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1696}{121}\quad \text{(upper right particle)}\\ r_3^2 &= \Big(\frac{24}{11}\Big)^2 ...


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Okay, so I figured it out myself. Here's what I think: Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular ...


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As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the ...


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I would suggest that the major reason for the ball becoming stationary is its inevitable interaction with the air - i.e. friction, resistance to being parted and eddy currents.



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