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I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ...


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The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem. We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is ...


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There is no such thing as "rotational speed". There is angular momentum, which has units of $\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}}$ (and so is not a speed), and angular velocity which has units of $\frac{1}{\mathrm s}$ (and so is not a speed either). There is no maximum angular velocity in special relativity. So long as $\omega r\lt c$ (the ...


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For high-performance graphics, where this thing has to be done all the time, the most common way to do these sorts of rotations is to store quaternions rather than Euler angles. You can convert between them with these techniques and then use them do to spatial rotations, Quaternions are not the simplest way to go but realistically, it's probably not going ...


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Equation 24 of your paper says: $$\vec{u}=\vec{U}+\vec{\omega}\times\vec{R}$$ Here $\vec{U}$ is the position of the center of mass, $\vec{\omega}$ is the angular velocity vector, and $\vec{R}$ is the position of one point in the rigid body. Taking the derivative, ...


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In polar coordinates you have $(x,y) = (r \cos \theta, r \sin \theta)$ Taking total derivatives of the above one finds that: Positions $$\begin{aligned} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \begin{pmatrix} r \\ 0 \end{pmatrix} \end{aligned} $$ Velocities ...


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Does gravity play a role in the Earth's equatorial bulge? Absolutely. A metal sphere: A very rigid object. If this is rotated then there is no bulge because it requires a lot of energy to deform the object. Of your three models, this is the closest to the truth. Metal is not quite as rigid as you think. It compresses under pressure, bends under ...



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