Question Number 214742 by mr W last updated on 18/Dec/24
Commented by mr W last updated on 18/Dec/24
$${find}\:{the}\:{volume}\:{of}\:{water}\:{in}\:{the} \\ $$$${cylinderical}\:{cup}. \\ $$
Commented by Hamada1969 last updated on 18/Dec/24
Commented by ajfour last updated on 18/Dec/24
https://youtu.be/pAkC2541W4Q?si=Pn7NfbYt0vSZwa-5
A video lecture of a projectile motion question by me.
Answered by ajfour last updated on 18/Dec/24
Commented by ajfour last updated on 18/Dec/24
$$\frac{{x}}{{L}}=\frac{{y}}{{R}} \\ $$$${y}=\left(\frac{{R}}{{L}}\right){x}\:\:\:\: \\ $$$$\mathrm{cos}\:\theta=\frac{{R}−{y}}{{R}}=\mathrm{1}−\frac{{x}}{{L}} \\ $$$${x}=\mathrm{0}\:\:\Leftrightarrow\:\theta=\mathrm{0}\:\:\:,\:{x}={L}\:\Leftrightarrow\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${dx}={L}\mathrm{sin}\:\theta \\ $$$$\int{dV}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left({R}^{\mathrm{2}} \theta−{R}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta\right)\left({L}\mathrm{sin}\:\theta\right) \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \theta\mathrm{sin}\:\theta{d}\theta=\left(−\theta\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{{V}}{{R}^{\mathrm{2}} {L}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{{V}}{\pi{R}^{\mathrm{2}} {L}}=\frac{\mathrm{2}}{\mathrm{3}\pi}\:\approx\:\mathrm{0}.\mathrm{2122} \\ $$$$ \\ $$