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I would approach this problem by "thinking" that the gear $A$ is like on a treadmill. So the gear $O$ is like moving "backward" while the gear $A$ is trying to move forward. You can easily compute the velocity of the two contact points and see what happens.

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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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Electron spin is not the result of a rotation of the electron around itself. In this case, of course (2) also doesn't hold. In fact, one can show that there is a double implication as follows: 1) if $\vec v$ is defined as in (1) one gets $$\frac {\vec r \times \vec v}{r^2} = \vec {\omega} - \vec r \frac {(\vec r \cdot \vec {\omega})}{r^2}. \tag{I}$$ So, ...

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