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Question Number 20375 by Tinkutara last updated on 26/Aug/17

$$\mathrm{A}\mathrm{small}\mathrm{particle}\mathrm{of}\mathrm{mass}m\mathrm{is}\mathrm{projected}$$$$\mathrm{at}\mathrm{an}\mathrm{angle}\theta \mathrm{with}\mathrm{the}x-\mathrm{axis}\mathrm{with}\mathrm{an}$$$$\mathrm{initial}\mathrm{velocity}{v}_0\mathrm{in}\mathrm{the}x-y\mathrm{plane}\mathrm{as}$$$$\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{Figure}.\mathrm{At}\mathrm{a}\mathrm{time}$$$$t<\frac{v_0\sin \theta }{g},\mathrm{the}\mathrm{angular}\mathrm{momentum}\mathrm{of}$$$$\mathrm{the}\mathrm{particle}\mathrm{is}$$

Commented by Tinkutara last updated on 26/Aug/17

Answered by ajfour last updated on 26/Aug/17

$$\overline{L}={\int }_0^t(x\hat{i}+y\hat{j})\times (-mg\hat{j})dt$$$$=-{\int }_0^t\left[\left(v_0\cos \theta \right)\left(mg\right)\hat{k}\right]tdt$$$$=-\left({mgv}_0\cos \theta \right)\frac{t^2}{2}\hat{k}$$$$\mid \overline{L}\mid =\frac{1}{2}\left({mgv}_0\cos \theta \right)t^2.$$

Commented by Tinkutara last updated on 26/Aug/17

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

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