Question

Expand the triple product for $\hvec{a}=\omega\times(\omega\timesr).$ . if r is perpendicular to ω , show that $\hvec{a}=-\omega^2\hvec{r}$  and so find the elementary result that the acceleration is toward the center of the circle of magnitude $\frac{\upsilon^2}{r}$

Answer 1

$$\bold{\omega \times(\omega \times r)=(\omega\cdot r)\omega-(\omega\cdot\omega)r}$$

$$(\omega\cdot r)=0$$

$$\bold{\omega \times(\omega \times r)=-\omega^2r}$$

$$\bold{a=-\omega^2r=-e_r}\frac{v^2}{r}$$