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4

Throughout the entire process, if we consider the system comprising both the ball and the ring, the external forces acting on the entire system would be the weight of the ball, the weight of the ring, and the normal force from the ground on the ring. Notice that none of these have any horizontal components. Thus, the horizontal position of the system's ...


0

The angular displacement is 3(2π) – π/3 radians (unless you assume one revolution gives a zero displacement). The final position is π/2– π/3 radians (It starts on the y axis).


0

Angular displacement is how much angle was changed in certain process. If angle changes by $+\pi$, then $+\pi$, and $+\pi$ again, then angular displacement is $3\pi$. Angular position, on the other hand, does not depend on process. That is what you see and measure when you see the system. Since the angle is cyclic, angular position is remainder of angular ...


0

The equation is always $$ \boxed{\; \boldsymbol{v} = \boldsymbol{\omega} \times \boldsymbol{r} \; } $$ where $\boldsymbol{\omega} = \pmatrix{\omega_x \\ \omega_y \\ \omega_z } $ and $\boldsymbol{r} = \pmatrix{ r \cos \phi \sin \theta \\ r \sin \phi \sin \theta \\ r \cos \theta} $ In your first case $\omega_z \neq 0$ while others are 0, and in the second ...


1

Consider a level of the second type with load $m$ at distance $r$ and a force perpendicular to a lever acting at a distance $d$. The system is in zero gravity. If the lever rotated by a small angle $\Delta\phi$, then the force has done the work $W=F\Delta s=Fd\Delta\phi$. On the other hand this energy was spent to increase the kinetic energy of the load: $$ \...


1

Assuming you're right, a little index relabelling helps:$$\begin{align}A_i^\prime&=A_i-\epsilon_{mlk}d\phi_l r_k\partial_mA_i+d\phi_k\epsilon_{ikj}A_j\\&=(\delta_{ij}(1-\epsilon_{mlk}d\phi_lr_k\partial_m)+d\phi_k\epsilon_{ikj})A_j.\end{align}$$In these three terms, the second (third) has been relabelled with $j\leftrightarrow m$ ($j\leftrightarrow k$)...


0

In a static situation, where nothing is accelerating, the sum of torques about any point (or axis) must be zero. In an isolated system of movable masses, the sum of torques about a chosen point will determine the rate of change of the angular momentum about that point.


0

If I undertsand correctly, you are asking how moment of inertia is same for some objects about different axes. Moment of inertia particularly depends on distribution of mass from the axes $$I=\int r^2dm$$ Where, $r$ is perpendicular distance of small mass $dm$ from the given axes. For moment of inertia about different axes to be same, the distribution of ...


1

For a pivoted body the following relationship holds $$ \tau = I \dot{\omega} + \tau_w $$ where $\tau$ is the torque at the pivot, $I$ is the mass moment of inertia about the pivot (equals $\tfrac{m}{3} \ell^2$ for a rod), $\dot{\omega}$ is the rotational acceleration, and $\tau_w$ is the torque caused by external forces like gravity. This is something like $\...


2

A perfectly incompressible ball that experiences no air friction is an approximation to reality. This approximation disregards friction. So you should not expect it to give you realistic results about friction. If you disregard all sources of friction and other loss, you would find the ball rolls forever.


2

Note that Weinberg describes $R^{-1}(\mathcal{R}\hat{\textbf{p}}) \mathcal{R} R(\hat{\textbf{p}}) $ and not $R(\mathcal{R}\hat{\textbf{p}}) \mathcal{R} R^{-1}(\hat{\textbf{p}}) $. $R(\hat{\textbf{p}})$ rotates the $z$-axis into the direction of $\hat{\textbf{p}}$, this then gets rotated into a new direction $\mathcal{R}\hat{\textbf{p}}$ and finally $R^{-1}...


-1

Yes. The spin part of the wave function is $$e^{im_z\phi} \,.$$ It is invariant under $2\pi/m_z$ rotations about the quantization axis. The rotation angle depends on $m_z $ rather than on $S$. For a quartet state with $m_z=\pm 3/2$ this implies invariance under a (multiple of) $4\pi/3=240^\text{o}$ rotation(s). This includes $4\pi$ rotations. Quartet or any ...


7

No... Not if you are asking about the period of the rotation matrix in the quartet representation. You know that for the quartet representation, $j=3/2$, rotations, the group element is a 4×4 rotation matrix, $$ e^{i \theta (\hat{\boldsymbol{n}}\cdot\boldsymbol{J})} = I_4 \cos (\theta/2)\left(1+\tfrac{1}{2}\sin^2 (\theta/2)\right)+(2i \hat{\boldsymbol{n}}\...


0

I feel that my machinery is close. When the entirety of the information regarding the rotating and stationary frames is completely determined by the vector addition of position vectors, why on earth does one need to introduce this bizarre notation of fixed/rotating in the derivatives? Your machinery is not close. What you are missing is a concept that ...


2

To express the equation of motion in the non inertial system you have to take the first and second derivatives of position in the non inertial system. You relate the second derivative (acceleration) in the non inertial system to the second derivative in the inertial system to obtain the fictitious forces in the non inertial system. I do not know of a valid ...


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