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5

For a single particle, yes they're parallel. For a system of particles, $$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$ (you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted ...


0

You don't need angular velocity to find the linear (translational) speed. If you have answered question d.ii then you have an equation like $$T_{cen}=ma_{cen}$$ where $a_{cen}$ is the central (radial) acceleration, and $T_{cen}$ is central (radial) force towards the center. You now need to know the following formula for a circular orbit: ...


0

Use Fresnel coordinates: $$ ma_n=m\frac{v^2}{r}=F_n\\ ma_t=\frac{dv}{dt}=F_t\\ $$ Now $F_n$ denotes the total forces in the radial direction, and using a diagram you find: $$ F_n=P_n+T_n=T\sin{(30)} $$ Now for the tangential direction do the same thing to find: $$ F_t=P_t+T_t=-mg+T\cos{(30)} $$ Note that the speed is constant so $dv/dt=0$: $$ F_t=0\\ ...


1

While not necessary to solve this problem, I want you to know that... Concept # 0: the angular velocity of circular motion is directly proportional to the linear velocity of motion, $$ v = \omega r $$ where $v$ is the linear velocity, $\omega$ is the angular frequency, and $r$ is the radius of circular motion. Concept # 1: whenever an object exhibits ...


0

Why is your answer incorrect? Because at first you're calculating velocity of $C$ assuming point $A$ is at rest. Then the velocity you find for point $O$ is its velocity relative to the point $A$. Similarly, when you're calculating the velocity of point $A$ what you find is actually its velocity relative to $O$. They are the same thing, only with a sign ...


0

You must remember that the 'body-frame' (which is the frame 'inside' the rotating body) is a frame and that OTHER bodies (which may, for example, be rotating about the same point with a different speed) will appear in the 'body-frame' to have an angular speed less than if observed in an inertial (stationary) frame. To help visualise, if you stick your arm ...


0

Lets put the observer at A and the particle at B. The position kinematics are: $\vec{r}_A(t) = (a,0,0)$ $\vec{r}_B(t) = (r \cos \phi,r \sin \phi,0)$ The velocity kinematics are $\vec{v}_A(t) = (0,0)$ $\vec{v}_B(t) = (-r \omega \sin \phi,r \omega \cos \phi)$ The acceleration kinematics are $\vec{a}_A(t) = (0,0)$ $\vec{a}_B(t) = (-r \omega^2 \cos ...


0

I think the distinction here is between a free vector and a line vector. A free vector is shared by the entire rigid body regardless of location. Examples are: angular velocity $\mathbf{\omega}$, force $\mathbf{F}$ linear momentum $\mathbf{p}$. line vector is defined to act on a specific location (for example a point A), and along a specific ...


3

I don't think this is as dumb a question as everyone downvoting you seems to think. The definitions of angular velocity ($\omega$), angular momentum ($L$), and moment of inertia ($I$) ARE defined in order to perfectly mirror Newton's laws. Angular momentum is the analogue of momentum, angular velocity is the analogue of velocity, and moment of inertia is ...


2

Yes, see "Measurements of length of day [LOD] using the Global Positioning System" (its persistent Digital Object Identifier is doi:10.1029/96JB01889) and subsequent citing literature. Notice this is not a relativistic effect. LOD is day-averaged measure of Earth's spinning rate, see Wikipedia for background (LOD variations).


1

$v = |\vec{\omega}\times\vec{r}|= \omega r$ is always valid for a rigid rotating body. Here, $r$ refers to the distance of any particular point from a chosen axis of rotation, $\omega$, the angular speed of the body about that chosen axis and $v$, the linear speed of that point perpendicular to the radius vector (or the line joining the axis to that ...


0

As Alfred Centauri mentioned, the ring is indeed slipping. Think about a ring whose center is fixed but is at the same time rotating. Clearly $v=0$ but $\omega\neq 0$. In general, a well defined relationship between translational velocity $v$ and rotational velocity $\omega$ exists in the case of a ring that is rolling without slipping. Othewise, there is ...


3

To a first approximation it would make the longitude of the ground station shift. Assuming the satellites are unaffected by whatever slowed the Earth (ie. tidal forces or seismic activity) But the only way to get a GPS position sufficiently accurate to see the effect would be to use a differential GPS system which compares the incoming satellite signal to ...



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