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I suppose that in class you were shown how the concept of moment of inertia comes about, where the motion of a rigid body is divided into the motion of center of the mass and the motion about the center of mass. If so, you can see how the first part in your expression for the kinetic energy ($\frac{1}{2}mv_{CM}^2$, with emphasis on the $CM$ in the subscript)...
Angular momentum transforms according to the following rule: $$\vec{L}_A = \vec{L}_C + \vec{p} \times \vec{r}_{A/C}$$ which is similar to the rule for velocities $$\vec{v}_A = \vec{v}_C + \vec{\omega} \times \vec{r}_{A/C}$$ where $\vec{r}_{A/C} = \vec{r}_A - \vec{r}_C$ is the relative location of A as seen from C. So you start with $\vec{L}_C = I_C \... 5 This answer is probably too advanced for the OP, but might be interesting to those who have studied more advanced math or physics. (I mean this entirely respectfully.) The "best" way to mathematically represent a (proper) rotation in 3D space is not by a vector, but instead by an element of the Lie group$\mathrm{SO}(3)$.$\mathrm{SO}(3)$is a three-... 11 There's two possible views here. One of these is that you can, indeed, consider angular displacement or position as a vector in that it can be encoded with one: if you have $$\mathbf{\Theta} := \left<\theta_x, \theta_y, \theta_z\right>$$ then you can declare that this encodes an angular displacement with axis$\mathbf{\hat{\Theta}}$and with ... 9 The key is in the parenthetical statement in your first block quote: focus on the "unless they are very small" part. This can be seen by doing the simple "experiment" below. While this answer is not mathematically rigorous (for the rigor, see @The_Sympathizer's answer), I think it gets at the heart of the idea behind how we can get angular velocity as a ... 4 Vectors are mathematical expressions which should transform in the right way. Vector transformation means the way common vectors (like displacement) transforms under the translational or rotation of coordinate system. Can a vector arise from a non-vector quantity? Yes and they always do. Gradient of any scalar field is a vector field. There is also ... 0 Angular displacement has to specify a plane. Then it is a vector as long as you only add it up with corresponding vectors( of the same plane). 3 Angular position can be considered a vector$\vec \theta$. Just not a so-called geometric vector (or Euclidean vector). The author of your text book implicitly refers to geometric vectors, it seems. Many mathematical operations are defined for geometric vectors only, since we here see high degrees of symmetry and thus see simple relations such as the ... 0 Roll a wheel down the road. If it’s really rolling, the place where the wheel touches the road doesn’t move relative to the road: it’s just sets down as the wheel gets to a spot, then lifts up after. On the other hand, the top of the wheel is moving twice as fast as the axle. The skyhook is like that. It’s engineered to be rotating and orbiting at just the ... -2 I have come to realise that no direct relationship such as$\omega_x = \frac{d \theta_x}{dt}$emerges out of the definition for angular velocity$\overrightarrow {\omega} =\widehat{n} \frac{d \theta}{dt}$. If you see the first picture, you will notice that${\omega_x}$is 0, since$\overrightarrow {\omega}$points in$\widehat{k}$direction. However, in ... 1$v=\dot x$is the rate of change of extension of the spring, which equals the difference in velocities between its 2 ends. I would assume that the spring is massless. As soon as the ball touches one end of the spring the compression force starts to act on the trolley at the other end. The spring is not attached to the ball so the ball loses contact again ... 0 Your approach is correct. You recognized there is only one degree of freemdom, and you have chosen$x$to be it. The rotation$\theta$of the ball is dependent on$x$given a no-slip condition. With a simple free body diagram, you formulate the equation of motions. The forces acting on the ball are the spring force and the friction to the ground. $$\begin{... 0 Regardless of what the nature of the recorded data is, all vectors transform the same. In other words, it doesn't matter whether its \vec{a} or \vec{\omega} as long as its a vector quantity. Whatever you used to correct the linear accelerations, \vec{a}, can exactly be used for \vec{\omega} as long as the orientation of the sensor was fixed during ... 1 If the small disk did not rotate relative to the large disk then its rotation rate (angular velocity) relative to us would be the same as that of the large disk \omega_0. This answer does not depend on the distance x between the centres of the two disks. If the small disk rotates at rate \omega_1 relative to the large disk then its rotation rate ... 1 For a formal proof, remember that in 2D$$\hat{r} = \cos \theta \cdot \hat{x} + \sin \theta\cdot \hat{y}$$Because the object you're pointing at is moving, its direction vector has a derivative:$$ \frac{d}{dt} \hat{r} = (-\sin \theta)\cdot \dot\theta \cdot \hat x + (\cos \theta)\cdot \dot\theta \cdot \hat y = \dot\theta \cdot (-\sin \theta \cdot \hat{x}... 2 Imagine the scenario where you have an object (let’s say a coin) on a revolving platform. The first term is simply the velocity of the coin as it moves across the platform, irrespective of the platform rotating. In the case where the radius is fixed, this term is zero as$\dot{r}=0\$ for a constant radius. There’s certainly more rigorous ways of deriving ...