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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

$$\overrightarrow{L}=m\overrightarrow{r}\times \overrightarrow{v}$$$$=m\left(r\hat{r}\right)\times \left(\omega r\hat{\theta }\right)$$$$=m\omega r^2\hat{z}$$$$m,\omega ,r\mathrm{are}\mathrm{all}\mathrm{constants}\mathrm{and}\mathrm{the}\mathrm{vector}$$$$\mathrm{is}\mathrm{always}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{direction}\hat{z}.$$

Commented by Tinkutara last updated on 20/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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