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}"
"(\\omega\\cdot r)=0"
"\\bold{\\omega \\times(\\omega \\times r)=-\\omega^2r}"
"\\bold{a=-\\omega^2r=-e_r}\\frac{v^2}{r}"
Comments
Leave a comment