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

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

Fig. P15.137

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,

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

where $\alpha =\alpha \text{k}$ is the angular acceleration of the slab.