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I think you are asking about how to transform various vector quantities between points attached to a rigid body. Here are a rundown of the rules of transformation between an arbitrary point A (located at $r_A$) and the center of mass C (located at $r_C$). \begin{align} v_C & = v_A + \omega \times (r_C - r_A) & & \text{linear velocity at C} \\ ...

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The force of friction acts both towards the centre of the circle and opposite the velocity vector of the car. Strictly speaking, the diagram you have does not show all forces acting on the car but it is enough for purposes of explaining the circular motion. As the text also explains, circular motion always requires a force pointed radially inwards because ...

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If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration. In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.

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does it also leads to ... No, it doesn't. A simple Counterexample: Consider the figure below (the bar $\textrm {AB}$ is on a plane parallel to $\textrm {xy}$ plane) We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$ is the ...

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