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The casing will spin in the opposite direction. That is the principle of reaction wheels.


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I would do a hand analysis at first. You have two torques on the arrow. One is the drag torque, which is maximum when the arrow is transverse to the orbital velocity, as in your initial condition. The second is gravity gradient torque, which will be maximum when the arrow is horizontal. Compute each of these for your arrow. If one is much greater than ...


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Oh, absolutely. With a Superman-like body you can also 1) Lift an entire mountain in two hands without the rock disintegrating under the local pressure, and without the mountain falling apart due to preexisting weaknesses in the rock, 2) Travel faster than light 3) Travel faster than light and travel backwards in time 4) Hear sounds so faint that they ...


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I believe you could overcome gravity (in the atmosphere) in the same way as birds do - you just need enough muscles to do that.


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Your entire analysis is entirely correct and complete (and in my opinion the best way to go about it). There's no more information that need to be derived about this issue. Other analyses are just a different representations of the same facts.


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I am going describe the experiment that your question concerns, then I will restate your question, and then I will resolve your question and explain what you did wrong and how you could avoid make similar mistakes in the future. The experiment The experiment in question deals with rotational motion. The purpose of the experiment is to measure the ...


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By definition, both torque and angular momentum are relative to a point. So when you say $\tau=r \times f$ and $L=r \times p$, the vector $r$ originates from some fixed point. In light of these definitions, you can also easily prove that $dL/dt = \tau$ and various other useful properties. In classical mechanics, these are always true. One very useful ...


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Torque is the angular force (the moment) - i.e. $\tau = r \times F$, not $r \times p$, which is angular momentum. When you are dealing with a rotating or rotational system (like a disc on an axis), then when the torque and angular momentum are always aligned along that axis, it is much easier to use the magnitude of the torque and the magnitude of the ...


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The sum of torques as seen from any point relative to the object, which isn't changing its angular velocity, should add up to zero. In order to solve for all reaction forces for such a static equilibrium you have to pick the same number of points as you have mounting points. When you are calculating things by hand it is often easier to pick each point on ...


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Since it's in rotational equilibrium, the torques will sum to zero about any axis you choose. Sometimes there is an obvious choice of axis because you can ignore any forces that pass directly through the axis and using that one will make the analysis simpler. But doing the analysis about some other axis is exactly as valid.


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It is 1 mm high and has a "residual magnetism" of 1.4 T. I think that might be enough to estimate the dipole moment as we know that the magnetic field along the axis of a dipole goes as $$B = \frac{\mu_0}{2\pi r^3}\vec{m}$$ Putting $r = 1$ mm, $\mu_0 = 4\pi\cdot 10^{-7} $H/m, and $B = 1.4$ T, we find $$m = \frac{2\pi r^3 B}{\mu_0} = 7 \cdot ...


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Suppose you pedal at about 100 rpm (I don't know what a typical rate of pedalling is, but this seems a plausible order of magnitude). To make a fair comparison with your motorbike you need to gear the moorbike engine down from 7,000 rpm to the same 100 rpm that you pedal at, i.e. a factor of 70, and this will multiply the torque by a factor of 70. So at 100 ...


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You're using a vector formula with cross product. Let's switch to this: t = r * F * sin theta, where theta is the angle between your leg and the lever on the sprocket wheel. Theta will change depending on where you are in the cycle of pedaling your bike. Greatest value for sin theta will be when the sprocket crank is horizontal and your leg is ...



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