Question Number 29687 by ajfour last updated on 11/Feb/18
Commented by ajfour last updated on 11/Feb/18
$${Find}\:{the}\:{final}\:{velocity}\:{gained}\:{by}\:{a} \\ $$$${tank}\:{of}\:{mass}\:{M},\:{initially}\:{completely} \\ $$$${filled}\:{with}\:{water}\:{and}\:{at}\:{rest},\:{as}\:{it} \\ $$$${slips}\:{on}\:{a}\:{frictionless}\:{ground}\:{owing} \\ $$$${to}\:{a}\:{small}\:{opening}\:{at}\:{its}\:{base}\:{of} \\ $$$${radius}\:{a},\:{through}\:{which}\:{it}\:{drains} \\ $$$${out}.\:\left({Assume}\:{density}\:{of}\:{water}\:\boldsymbol{\rho}\right). \\ $$
Commented by 803jaideep@gmail.com last updated on 11/Feb/18
$$\mathrm{isnt}\:\mathrm{area}\:\mathrm{of}\:\mathrm{hole}\:\mathrm{given}? \\ $$
Commented by ajfour last updated on 11/Feb/18
$${yes}\:\pi{a}^{\mathrm{2}} \:\:\left({please}\:{read}\:{question}\:{again}\right). \\ $$
Commented by ajfour last updated on 11/Feb/18
$${whats}\:{the}\:{final}\:{speed}\:{of}\:{tank} \\ $$$${by}\:{the}\:{time}\:{y}\rightarrow\:\mathrm{0}\:. \\ $$
Answered by mrW2 last updated on 12/Feb/18
![M_W =mass of water at begin=ΟR^2 hΟ at water depth y: total mass: m=M+M_W (y/h) β(dm/dt)=(M_W /h)Γ(dy/dt) (relative) speed of outgoing water: v=(β(2gy)) letβ²s say speed of tank=u βv(dm/dt)=m(du/dt) βvΓ(M_W /h)Γ(dy/dt)=m(du/dy)Γ(dy/dt) β(M_W /h)v=m(du/dy) β(M_W /h)(β(2gy))=(M+(M_W /h)y)(du/dy) β(M_W /h)Γ((β(2gy))/(M+(M_W /h)y)) dy=du β(β(2g))Γ((βy)/((((hM)/M_W )+y))) dy=du with k=((hM)/M_W ) β(β(2g))Γβ«_h ^( 0) ((βy)/((k+y))) dy=β«_0 ^( u_(max) ) du (β(2g))Γ2(βk)[(β(y/k))βtan^(β1) (β(y/k))]_0 ^h =u_(max) u_(max) =2(β(2gk))((β(h/k))βtan^(β1) (β(h/k))) u_(max) =2(β((2ghM)/M_W ))((β(M_W /M))βtan^(β1) (β(M_W /M))) with Ξ»=(M_W /M) βu_(max) =2(β((2gh)/Ξ»))((βΞ»)βtan^(β1) (βΞ»)) Example: Ξ»=(M_W /M)=5, h=1 m βu_(max) =2(β((20)/5))((β5)βtan^(β1) (β5))=4.3 m/s ==================== u(y)=2(β(2gk))((β(h/k))β(β(y/k))βtan^(β1) (β(h/k))+tan^(β1) (β(y/k))) Οa^2 vdt=βΟR^2 dy β(dy/dt)=β(a^2 /R^2 )v=β((a^2 (β(2gy)))/R^2 ) β(dy/(βy))=β((a^2 (β(2g)))/R^2 ) dt ββ«_h ^( y) (dy/(βy))=β((a^2 (β(2g)))/R^2 ) β«_0 ^( t) dt β2((βy)β(βh))=β((a^2 (β(2g)))/R^2 )Γt β(βy)=(βh)β((a^2 (β(2g)))/(2R^2 ))Γt=(βh)βpt βy=hβ2(βh)pt+p^2 t^2 y=0βt=((βh)/p)=(R^2 /a^2 )(β((2h)/g))=t_1 (at t_1 tank is empty) βu(t)=2(β(2gk))[((pt)/(βk))βtan^(β1) (β(h/k))+tan^(β1) ((β(h/k))β((pt)/(βk)))] with Ξ·=(p/(βk))=(a^2 /R^2 )(β((2gΞ»)/h)) βu(t)=2(β((2gh)/Ξ»))[Ξ·tβtan^(β1) (βΞ»)+tan^(β1) ((βΞ»)βΞ·t)]](Q29759.png)
$${M}_{{W}} ={mass}\:{of}\:{water}\:{at}\:{begin}=\pi{R}^{\mathrm{2}} {h}\rho \\ $$$${at}\:{water}\:{depth}\:{y}: \\ $$$${total}\:{mass}:\:{m}={M}+{M}_{{W}} \frac{{y}}{{h}} \\ $$$$\Rightarrow\frac{{dm}}{{dt}}=\frac{{M}_{{W}} }{{h}}Γ\frac{{dy}}{{dt}} \\ $$$$\left({relative}\right)\:{speed}\:{of}\:{outgoing}\:{water}: \\ $$$${v}=\sqrt{\mathrm{2}{gy}} \\ $$$$ \\ $$$${let}'{s}\:{say}\:{speed}\:{of}\:{tank}={u} \\ $$$$β{v}\frac{{dm}}{{dt}}={m}\frac{{du}}{{dt}} \\ $$$$β{v}Γ\frac{{M}_{{W}} }{{h}}Γ\frac{{dy}}{{dt}}={m}\frac{{du}}{{dy}}Γ\frac{{dy}}{{dt}} \\ $$$$β\frac{{M}_{{W}} }{{h}}{v}={m}\frac{{du}}{{dy}} \\ $$$$β\frac{{M}_{{W}} }{{h}}\sqrt{\mathrm{2}{gy}}=\left({M}+\frac{{M}_{{W}} }{{h}}{y}\right)\frac{{du}}{{dy}} \\ $$$$β\frac{{M}_{{W}} }{{h}}Γ\frac{\sqrt{\mathrm{2}{gy}}}{{M}+\frac{{M}_{{W}} }{{h}}{y}}\:{dy}={du} \\ $$$$β\sqrt{\mathrm{2}{g}}Γ\frac{\sqrt{{y}}}{\left(\frac{{hM}}{{M}_{{W}} }+{y}\right)}\:{dy}={du} \\ $$$${with}\:{k}=\frac{{hM}}{{M}_{{W}} } \\ $$$$β\sqrt{\mathrm{2}{g}}Γ\int_{{h}} ^{\:\mathrm{0}} \frac{\sqrt{{y}}}{\left({k}+{y}\right)}\:{dy}=\int_{\mathrm{0}} ^{\:{u}_{{max}} } {du} \\ $$$$\sqrt{\mathrm{2}{g}}Γ\mathrm{2}\sqrt{{k}}\left[\sqrt{\frac{{y}}{{k}}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{y}}{{k}}}\right]_{\mathrm{0}} ^{{h}} ={u}_{{max}} \\ $$$${u}_{{max}} =\mathrm{2}\sqrt{\mathrm{2}{gk}}\left(\sqrt{\frac{{h}}{{k}}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{h}}{{k}}}\right) \\ $$$${u}_{{max}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{ghM}}{{M}_{{W}} }}\left(\sqrt{\frac{{M}_{{W}} }{{M}}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{M}_{{W}} }{{M}}}\right) \\ $$$${with}\:\lambda=\frac{{M}_{{W}} }{{M}} \\ $$$$\Rightarrow{u}_{{max}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{gh}}{\lambda}}\left(\sqrt{\lambda}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\lambda}\right) \\ $$$$ \\ $$$${Example}:\:\lambda=\frac{{M}_{{W}} }{{M}}=\mathrm{5},\:{h}=\mathrm{1}\:{m} \\ $$$$\Rightarrow{u}_{{max}} =\mathrm{2}\sqrt{\frac{\mathrm{20}}{\mathrm{5}}}\left(\sqrt{\mathrm{5}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\mathrm{5}}\right)=\mathrm{4}.\mathrm{3}\:{m}/{s} \\ $$$$==================== \\ $$$${u}\left({y}\right)=\mathrm{2}\sqrt{\mathrm{2}{gk}}\left(\sqrt{\frac{{h}}{{k}}}β\sqrt{\frac{{y}}{{k}}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{h}}{{k}}}+\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{y}}{{k}}}\right) \\ $$$$\pi{a}^{\mathrm{2}} {vdt}=β\pi{R}^{\mathrm{2}} {dy} \\ $$$$\Rightarrow\frac{{dy}}{{dt}}=β\frac{{a}^{\mathrm{2}} }{{R}^{\mathrm{2}} }{v}=β\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{2}{gy}}}{{R}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{\sqrt{{y}}}=β\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{{R}^{\mathrm{2}} }\:{dt} \\ $$$$\Rightarrow\int_{{h}} ^{\:{y}} \frac{{dy}}{\sqrt{{y}}}=β\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{{R}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:{t}} {dt} \\ $$$$\Rightarrow\mathrm{2}\left(\sqrt{{y}}β\sqrt{{h}}\right)=β\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{{R}^{\mathrm{2}} }Γ{t} \\ $$$$\Rightarrow\sqrt{{y}}=\sqrt{{h}}β\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{2}{g}}}{\mathrm{2}{R}^{\mathrm{2}} }Γ{t}=\sqrt{{h}}β{pt} \\ $$$$\Rightarrow{y}={h}β\mathrm{2}\sqrt{{h}}{pt}+{p}^{\mathrm{2}} {t}^{\mathrm{2}} \\ $$$${y}=\mathrm{0}\Rightarrow{t}=\frac{\sqrt{{h}}}{{p}}=\frac{{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\sqrt{\frac{\mathrm{2}{h}}{{g}}}={t}_{\mathrm{1}} \:\left({at}\:{t}_{\mathrm{1}} \:{tank}\:{is}\:{empty}\right) \\ $$$$ \\ $$$$\Rightarrow{u}\left({t}\right)=\mathrm{2}\sqrt{\mathrm{2}{gk}}\left[\frac{{pt}}{\sqrt{{k}}}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\frac{{h}}{{k}}}+\mathrm{tan}^{β\mathrm{1}} \left(\sqrt{\frac{{h}}{{k}}}β\frac{{pt}}{\sqrt{{k}}}\right)\right] \\ $$$${with}\:\eta=\frac{{p}}{\sqrt{{k}}}=\frac{{a}^{\mathrm{2}} }{{R}^{\mathrm{2}} }\sqrt{\frac{\mathrm{2}{g}\lambda}{{h}}} \\ $$$$\Rightarrow{u}\left({t}\right)=\mathrm{2}\sqrt{\frac{\mathrm{2}{gh}}{\lambda}}\left[\eta{t}β\mathrm{tan}^{β\mathrm{1}} \sqrt{\lambda}+\mathrm{tan}^{β\mathrm{1}} \left(\sqrt{\lambda}β\eta{t}\right)\right] \\ $$
Commented by mrW2 last updated on 12/Feb/18
Is any solution for this question given?
Commented by ajfour last updated on 12/Feb/18
$$\mathbb{SUPERB}!\:{Sir},\:{Energy}\:{method} \\ $$$${leads}\:{to}\:{a}\:{complicated}\:{differential} \\ $$$${equation},\:{i}\:{guess}.\: \\ $$$${Your}\:{method}\:{is}\:\mathscr{E}{xcellent}\:,\:{Sir}. \\ $$
Commented by ajfour last updated on 12/Feb/18
$${No}\:{sir},\:{not}\:{from}\:{book}. \\ $$