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If you want to take the sum of moments, you have to take **all the external forces ** into account. The summation of moments has no preference for a certain force, allowing it to be in the equation. All forces are in the summation, only their resulting moment can be zero. If we look at the following example, assuming that all the spacings, e.g. ...

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The proof is most easiest if we use the vector notation. We have $$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$ where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have, $${{\vec r}_{cm}} = \frac{1}{M}\int ... 1 There is a distinction between points and vectors. Points are positions in space, and vectors are directions. One can easily mix up the two, because in Euklidean space they look rather similar. \theta in this case is a coordinate, i.e. part of the description of a point. The vector associated to that coordinate could be called \hat{e}_\theta, and point ... 1 Equations of motion explained: Sum of forces on body equals mass times acceleration of center of mass.$$\sum \vec{F} = \frac{{\rm d} }{{\rm d}t}(m\vec{v}_{cm}) = m \vec{a}_{cm}$$Sum of moments on center of mass equals mass moment of inertia times angular acceleration of body, plus velocity related terms.$$\sum \vec{M}_{cm} = \frac{{\rm d} }{{\rm ...

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The same force applied anywhere on a rigid object causes the same translational accelleration. The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration. Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One ...

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A synchronous motor will not reduce in speed with increasing load until a certain load is reached. It will continue to run at the same speed, phased to the AC power line, until the torque demand exceeds what it can produce in that mode. At zero load, the voltage and current are out of phase, so the average electrical consumption is zero. As the load ...

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Hmmm, my relativity is weak, but... In the usual Newtonian formulation of physics forces are vector quantities which suggests that in the relativistic formulation they should be the spacial part of the Lorentz four-vector and should transform as such. Further looking at Newton's second law we see force expressed as a time-derivative of momentum, which is to ...

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Well, since this a physics forum, here is the physical answer: Moment of Inertia if defined as such: $$\tau=I \alpha$$ Where $\tau$ is torque, $I$ is moment of inertia, and $\alpha$ is the angular acceleration. The moment of inertia for an evenly distributed "barrel" (A cylindrical honey-extractor, for example) is $$I=mr^2$$ You said $r=350mm, m=50kg$, ...

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