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5

The length of the arc PQ is $r\Delta\theta$, as Feynman says, but the difference from $\tan\Delta\theta$ or $2\tan (\Delta \theta/2)$ or something else is negligible because $\Delta \theta$ is assumed to be infinitesimal (infinitely small) in the argument, anyway. For that reason, both angles OPQ and OQP should be taken to be 90 degrees.


3

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


3

General remarks. That's right. Torque is defined as $\mathbf r\times\mathbf F$ where $\mathbf r$ is the position where the force is applied, and $\mathbf F$ is the force being applied. The so-called Law of the Lever can then be derived from the following fact (which itself can be derived from Newton's Laws) about systems of particles: The net external ...


1

If the object is not spinning, then moving it will take the same effort whether the mass is spread out, or all at one point. Imagine that you are on a roundabout, but you are sitting on a disk that it itself able to spin. Initially you are facing North, and you are exactly in the middle of your disk (although it is off axis of the roundabout). Now the ...


1

This is a very late response, but there is no accepted answer as of yet, and none of the answer quite hit the mark. Regarding the magical collision hypothesis, that smacks of being rather non-scientific. Scientists as well as Missourians are wont to say, "Show me!" Other than the fact that Venus's rotation is anomalous, what, exactly, is the evidence for ...


1

As others have already pointed out, the moment of inertia depends mostly on the mass distribution. As an example, imagine two cylinders with exactly the same mass. Suppose that you cannot see what is inside them, it's covered on both ends (and the covers mass is negligible). The first is a solid cylinder. The second one is a hollow cylinder. Ka-boom! ...


1

Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix $a$ that relates the two sets of axes is simply the identity matrix. $dG'$ is a vector, so $a_{ji}dG_j' = dG_i'$ is simply equivalent to the statement that with $I$ the identity matrix and $V$ an arbitrary vector, $IV=V$. Answer ...


1

If the sphere is not spinning, and the axis about which you are taking angular momentum stays through the same position in the sphere, then the relative velocity of the sphere to the axis will always be zero regardless of translational motion. Therefore, the angular momentum will be zero.



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