Chapter 15.4, Problem 15.137P

Denoting by r_A the position vector of a point A of a rigid slab that is in plane motion, show that (a) the position vector r_C of the instantaneous center of rotation is

$r_C=r_A+\frac{\omega \times v_A}{{\omega }^2}$

where ω is the angular velocity of the slab and v_A is the velocity of point A, (b) the acceleration of the instantaneous center of rotation is zero if, and only if,

$a_A=\frac{\alpha }{\omega }{v}_A+\omega \times v_A$

where α = αk is the angular acceleration of the slab.

Fig. P.15.137