Question Number 21295 by Tinkutara last updated on 19/Sep/17
$$\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{performing}\:\mathrm{uniform} \\ $$$$\mathrm{circular}\:\mathrm{motion},\:\mathrm{angular}\:\mathrm{momentum}\:\mathrm{is} \\ $$$$\mathrm{constant}\:\mathrm{in}\:\mathrm{magnitude}\:\mathrm{but}\:\mathrm{direction} \\ $$$$\mathrm{keeps}\:\mathrm{changing}. \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{or}\:\mathrm{wrong}? \\ $$
Answered by dioph last updated on 19/Sep/17
$$\overset{\rightarrow} {{L}}\:=\:{m}\overset{\rightarrow} {{r}}×\overset{\rightarrow} {{v}} \\ $$$$=\:{m}\left({r}\hat {{r}}\right)×\left(\omega{r}\hat {\theta}\right) \\ $$$$=\:{m}\omega{r}^{\mathrm{2}} \hat {{z}} \\ $$$${m},\:\omega,\:{r}\:\mathrm{are}\:\mathrm{all}\:\mathrm{constants}\:\mathrm{and}\:\mathrm{the}\:\mathrm{vector} \\ $$$$\mathrm{is}\:\mathrm{always}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\hat {{z}}. \\ $$
Commented by Tinkutara last updated on 20/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$