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I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$... 3 The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:$$D_t = d_t + \omega(t)\times\tag{1}$$where D_t is the "total" derivative, i.e. the time derivative in the inertial frame, d_t is the time derivative in the frame fixed to the body. Since there there are no torques ... 2 The diagram you need is this: Obviously, the horizontal position of the center of mass is centered between the two H atoms; the vertical position is the point where the molecule would balance. So if the distance from the H-H line to S is h, and the mass of the atoms is m_O and m_H, then balance happens when$$2 m_H h = m_O (b-h)$$From there you ... 2 To answer your questions, would a 90 degree shot turn the ship, in theory yes, due to the conservation of angular momentum law. As long as the cannon was not exactly amidships , in which case it would push, or more probably just cause a list, to port or starboard. And would a 180 degree shot (aftwards) increase the speed of the ship, again, in theory ... 1 You could use a little outside knowledge and observe that for any Lie group with lie algebra \mathfrak{g} with X\,Y\in\mathfrak{g}:$$\exp(Y)\,\exp(X)\,\exp(-Y) = \exp(Z);\\\\ Z= X + [Y,\,X] + \frac{1}{2!} [Y,\,[Y,\,X]]+\frac{1}{3!}[Y,\,[Y,\,[Y,\,X]]]+\cdots$$This is another form of the so called braiding formula \mathrm{Ad}(e^Y) = ... 1 For high-performance graphics, where this thing has to be done all the time, the most common way to do these sorts of rotations is to store quaternions rather than Euler angles. You can convert between them with these techniques and then use them do to spatial rotations, Quaternions are not the simplest way to go but realistically, it's probably not going ... 0 Yes, is a Taylor expansion! Take the rotational direction and define this direction as z of some cartesian coordinate system (you have freedom to do that, you are a free physicist). Ignore the z component of the vector, because don't change with the rotation. The rotation take the form of:$$ \vec{v}(\theta)=\left( \begin{array} \ v_x(\theta) \\ ...

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It is a Taylor series of the rotation matrix, but the idea is that we don't know the form of the matrix when we write the expansion. Instead we use the definition of rotations, as linear transformations which preserve the length of vectors, to determine the form of the expansion coefficients. Its not hard to prove that every rotation matrix is orthogonal, ...

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It is easy to verify that every proper rotation in space, which is the subgroup of the full group of rotations of $\mathbb R^3$ that is connected to the identity $I$, can be expressed in the form $$R_u = e^{A_u},$$ where $A$ is a $3\times3$ skew-symmetric matrix. Indeed $(e^A)^Te^A = e^{-A}e^A = I$ and $\det(e^A) = e^{Tr(A)} = 1$ and therefore $e^A$ is a ...

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The rotation group of three dimensional space has three generators $T^a$ given by  T^3 = \left(\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right) \quad T^2 = \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{matrix}\right) \quad T^1= \left(\begin{matrix} 0 & 0 & 0 \\ 0 ...

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I'm afraid I cannot quite understand the specific part of your question. How would the motion always be perpendicular to the string? That would only be true if the mass were moving in a circle, not a decreasing spiral like you are describing. But to your main question, does the conservation of angular momentum hold if the center of rotation changes.. For ...

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