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If I understand your question correctly you are saying that: $$ v = r\omega $$ and therefore: $$\begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} $$ but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...


3

In the reference frame of the car, the axle is stationary, but the ground is moving below at speed $v_C$. If the car doesn't skid, then the surface of the tyre must move at the same speed, but with a velocity that is directed backwards in the bottom and forwards in the top. At half the distance between the tyre surface and the hub, the speed is $v_C/2$. ...


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Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...


1

First, angular momentum isn't measured about an axis. It's measured about a point. Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in. Now, about ...


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A rotating reference frame is an accelerated reference frame so $A$ and $B$ are at rest in an accelerated reference frame. Assume an inertial reference frame $S_0$ and another reference frame $S$, with a common origin and rotating with respect to $S_0$. Let the (constant) angular velocity vector of $S$ be $\mathbf \Omega$. Then, the time rate of change of ...



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