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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 ...


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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 ...


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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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Moment of force is torque. The torque that an inertia provides will be proportional to it's angular acceleration. So you don't actually have enough information in your question. The angular acceleration needs to be known to tell you what torque that inertia will provide. Simply put, how fast ($\omega$) must the dc starter turn and how soon must it get to ...


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From the fat that the units are different you have probably concluded there isn't a single "correct" answer - because it depends on how you convert one to the other. When a flywheel (with a certain moment of inertia $I$) is rotating with an angular velocity $\omega$, it has an angular momentum $L=I\omega$. The torque this can produce depends on the rate at ...


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This question lacks one important value in order for the answer you are looking for and that is the RPM required to get the generator started. This is because you can get the world's biggest flywheel attached, but if it is moving at a snail's pace, your generator won't start anyway. Also the "starting inertia" of $0.522~kg/m²$ should really read as ...


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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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This sounds to me like an experimental question (but be careful not to hurt yourself!). Note that since you want the chair to rotate forwards towards the ground, you'll want to consider the direction of the angular momentum your motion introduces. If you kick your legs out rapidly, not only does the weight of your feet tilt the chair forward, but the ...


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A force acts on a line and picking any point along the line makes no difference in terms of the moment produced. In fact, the cross product is equivalent to the area of the triangle formed by the force vector and the point O by which the moment of calculated. In the area of the triangle it is the height of the triangle (perpendicular distance) that is ...


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Both scalar and vector products are definitions. $$ {\bf a} \cdot {\bf b} = a b \cos \theta\,\,$$ where $a$ and $b$ are the magnitudes of the two vectors and $\theta$ is the angle between them. This quantity only has a magnitude and hence no direction and is therefore a scalar. If you apply a (let's say constant for the sake of simplicity) vector force ...


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This is an issue of not specifying the initial and final states (or whether we are calculating the work done by the magnetic torque, or the work required by an external torque to oppose the magnetic torque). In reality, what we should say is that if we start with a magnetic dipole $\vec{\mu}$ that is aligned with the magnetic field (so that $\theta=0$), WE ...



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