Answer to Question #167192 in Physics for wadeeah

Question #167192

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}"


1
Expert's answer
2021-03-01T11:50:28-0500
"\\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}"


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