Say we have a rigid body $B$ and a particle $p$ in $B$. Let $F$ be a fixed inerial reference frame and let $F'$ be a (possibly non-inertial) frame embedded in $B$. On one hand, I know that the velocity of $p$ relative to $F$ is given by

$$v_p = v_o+ w \times (r-O')$$ where $v_o$ is the velocity of the "origin" O' of $F'$, $w$ is angular velocity of the body $B$, and $r$ and $b$ denote the position vectors of $O'$ and $p$ with respect to $F$. On the other hand, I also know that

$$v_p = V_o + \Omega \times (r-O') + v' $$ where $v'$ denotes the apparent velocity of $p$ as seen in the frame $F'$

Comparing these two equations gives $v' = 0$, so does this mean that the apparent velocity of any particle in a rigid body relative to a fixed point in the body is zero?


That's the definition of a rigid body. No particle in the body has a velocity relative to a coordinate system that is fixed to the body.

  • $\begingroup$ See - that wasn't so hard... and now I can give you an uptick :-) $\endgroup$ – Floris Dec 12 '16 at 15:06

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