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Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ... 3 I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ... 0 First Thought (probably not the fastest) Let us assume you have a vector space in$R^{3}$with a quaternion defined as: $$\mathbf{q} = q_{F}^{*} \ q_{B} \\ = a + b \hat{\mathbf{x}} + c \hat{\mathbf{y}} + d \hat{\mathbf{z}}$$ where$(a, b, c, d)$are the Euler parameters and$(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})$defines the reference ... 0 The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates$\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates$\{t', x', y', z'\} $by making an appropriate substitution. For example, ... 1 Does gravity play a role in the Earth's equatorial bulge? Absolutely. A metal sphere: A very rigid object. If this is rotated then there is no bulge because it requires a lot of energy to deform the object. Of your three models, this is the closest to the truth. Metal is not quite as rigid as you think. It compresses under pressure, bends under ... 0 The inside of the earth is still, partly liquid, source: Mantle So, it's very viscous, slow moving under very high pressure, but it's still liquid enough and the Earth's crust is really quite thin and not nearly rigid enough to resist forces from below. Taking the earth as a whole, I'm not sure there's a really good single single analogy. A metal ... 2 The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem. We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is ... 1 For high-performance graphics, where this thing has to be done all the time, the most common way to do these sorts of rotations is to store quaternions rather than Euler angles. You can convert between them with these techniques and then use them do to spatial rotations, Quaternions are not the simplest way to go but realistically, it's probably not going ... 2 There is no such thing as "rotational speed". There is angular momentum, which has units of$\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$(the ... 0 That's correct so long as the rotation axis passes through the centers of mass of both objects. In general, the moment of inertia about a fixed axis (the$z-axis, say) will be something like $$I = \int_\text{object} \rho (x^2 + y^2) \,dV$$ But if we can split this integral up into two disjoint volumes (a cylinder and a sphere, say), we will have $$I = ... 0 Your familiar equation of a = r\alpha + r\omega^2 are for circular motion with a constant radius, i.e. swinging a mass around on a string. As you change the angular velocity (\alpha is nonzero), the velocity of the mass in the direction of its motion (tangent to the circle of radius r) will change as well. In addition, we learn in any physics class ... 0 If I understand your problem right, you want to see why the terms are radial/tangential: Radial -r \dot{\theta}^2 is indeed the term you know from a uniform circular motion. As in that case (UCM) only the direction of the velocity changes, the acceleration is perpendicular to the velocity, thus in the radial direction. \ddot{r} if the radius is ... 1 In polar coordinates you have (x,y) = (r \cos \theta, r \sin \theta) Taking total derivatives of the above one finds that: Positions$$\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} $$Velocities ... 1 Equation 24 of your paper says:$$\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, ...