Question Number 20375 by Tinkutara last updated on 26/Aug/17
$$\mathrm{A}\:\mathrm{small}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{projected} \\ $$$$\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:{x}-\mathrm{axis}\:\mathrm{with}\:\mathrm{an} \\ $$$$\mathrm{initial}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{the}\:{x}-{y}\:\mathrm{plane}\:\mathrm{as} \\ $$$$\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Figure}.\:\mathrm{At}\:\mathrm{a}\:\mathrm{time} \\ $$$${t}\:<\:\frac{{v}_{\mathrm{0}} \:\mathrm{sin}\:\theta}{{g}},\:\mathrm{the}\:\mathrm{angular}\:\mathrm{momentum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 26/Aug/17
Answered by ajfour last updated on 26/Aug/17
![L^ =∫_0 ^( t) (xi^ +yj^ )×(−mgj^ )dt =−∫_0 ^( t) [(v_0 cos θ)(mg)k^ ]tdt =−(mgv_0 cos θ)(t^2 /2)k^ ∣L^ ∣=(1/2)(mgv_0 cos θ)t^2 .](https://www.tinkutara.com/question/Q20379.png)
$$\bar {{L}}=\int_{\mathrm{0}} ^{\:\:{t}} \left({x}\hat {{i}}+{y}\hat {{j}}\right)×\left(−{mg}\hat {{j}}\right){dt} \\ $$$$\:\:\:\:=−\int_{\mathrm{0}} ^{\:\:{t}} \left[\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta\right)\left({mg}\right)\hat {{k}}\right]{tdt} \\ $$$$\:\:\:=−\left({mgv}_{\mathrm{0}} \mathrm{cos}\:\theta\right)\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\hat {{k}}\: \\ $$$$\:\:\mid\bar {{L}}\mid=\frac{\mathrm{1}}{\mathrm{2}}\left({mgv}_{\mathrm{0}} \mathrm{cos}\:\theta\right){t}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 26/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$