Question Number 227011 by fantastic2 last updated on 25/Dec/25
$${in}\:{a}\:{cup}\:{filled}\:{with}\:{a}\:{liquid}\:{of}\:{density}\:\rho,{a} \\ $$$${bubble}\:{rests}\:{in}\:{the}\:{liquid}\:{in}\:{h}\:{deepness}. \\ $$$${radius}\:{initial}\:{of}\:{the}\:{bubble}={r}_{{i}} \\ $$$$\:{initial}\:{density}\:{of}\:{the}\:{bubble}=\sigma_{{i}} \\ $$$$\left(\:{mass\&temparature}\:{is}\:{constant}\right) \\ $$$${a}\:{small}\:{jerk}\:{is}\:{given}\:{to}\:{the}\:{cup}\:{and}\:{the} \\ $$$${bubble}\:{starts}\:{to}\:{go}\:{upwards}. \\ $$$${find}: \\ $$$$\left.\mathrm{1}\right){velocity}\:\&\:{acceleration}\:{of}\:{the}\:{bubble}\:{at}\:{time}\:{t}\:{after} \\ $$$${it}\:{starts}\:{moving} \\ $$$$\left.\mathrm{2}\right){force}\:{acting}\:{on}\:{the}\:{bubble}\:{at}\:{time}\:{t} \\ $$$${after}\:{it}\:{starts}\:{moving} \\ $$$$ \\ $$
Commented by mr W last updated on 26/Dec/25
$${can}\:{you}\:{tell}\:{the}\:{source}\:{of}\:{the} \\ $$$${question}?\:{are}\:{you}\:{sure}\:{it}\:{is}\:{solvable}? \\ $$$${i}\:{mean}\:{can}\:{the}\:{time}\:{t}\:{be}\:{determined}? \\ $$
Commented by Kassista last updated on 27/Dec/25
Commented by Kassista last updated on 27/Dec/25
$${I}\:{got}\:{that}\:{result}\:{a}\mathrm{f}{ter}\:{some}\:{initial}\:{calculations},\:{but}\:{a}\:{final}\:{result} \\ $$$${considering}\:{the}\:{drag}\:{force}\:{exerted}\:{by}\:{the}\:{liquid}\:{would}\:{be}\:{too}\:{messy},\:{I}\:{believe} \\ $$
Answered by mr W last updated on 28/Dec/25
Commented by mr W last updated on 28/Dec/25
![assumed the size of the bubble is very small compared with the size of the tank. assumed the resistant force of the liquid is neglectable. m=V_i σ_i at depth h: bubble rests ⇒σ_i =ρ at depth x: (V/V_i )=((hρg+p_0 )/(xρg+p_0 ))=((h+h_0 )/(x+h_0 )) with h_0 =(p_0 /(ρg)) and p_0 =atmospheric pressure mg−Vρg=ma V_i σ_i g−(((h+h_0 )/(x+h_0 )))V_i ρg=V_i σ_i a g(1−((h+h_0 )/(x+h_0 )))=a ⇒a=g(1−((h+h_0 )/(x+h_0 ))) −v(dv/dx)=g(1−((h+h_0 )/(x+h_0 ))) ∫_0 ^v vdv=−g∫_h ^x (1−((h+h_0 )/(x+h_0 )))dx ∫_0 ^v vdv=−g∫_(h+h_0 ) ^(x+h_0 ) (1−((h+h_0 )/u))du (v^2 /2)=−g[(x−h)−(h+h_0 )ln ((x+h_0 )/(h+h_0 ))] ⇒v=(√(2g[(h−x)+(h+h_0 )ln ((x+h_0 )/(h+h_0 ))])) since the time t is infinite, we can not find a and v in terms of t.](https://www.tinkutara.com/question/Q227039.png)
$${assumed}\:{the}\:{size}\:{of}\:{the}\:{bubble}\:{is} \\ $$$${very}\:{small}\:{compared}\:{with}\:{the}\:{size} \\ $$$${of}\:{the}\:{tank}. \\ $$$${assumed}\:{the}\:{resistant}\:{force}\:{of}\:{the} \\ $$$${liquid}\:{is}\:{neglectable}. \\ $$$${m}={V}_{{i}} \sigma_{{i}} \\ $$$${at}\:{depth}\:{h}:\: \\ $$$${bubble}\:{rests}\:\Rightarrow\sigma_{{i}} =\rho \\ $$$${at}\:{depth}\:{x}: \\ $$$$\frac{{V}}{{V}_{{i}} }=\frac{{h}\rho{g}+{p}_{\mathrm{0}} }{{x}\rho{g}+{p}_{\mathrm{0}} }=\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} } \\ $$$${with}\:{h}_{\mathrm{0}} =\frac{{p}_{\mathrm{0}} }{\rho{g}}\:{and}\: \\ $$$${p}_{\mathrm{0}} ={atmospheric}\:{pressure} \\ $$$${mg}−{V}\rho{g}={ma} \\ $$$${V}_{{i}} \sigma_{{i}} {g}−\left(\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} }\right){V}_{{i}} \rho{g}={V}_{{i}} \sigma_{{i}} {a} \\ $$$${g}\left(\mathrm{1}−\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} }\right)={a} \\ $$$$\Rightarrow{a}={g}\left(\mathrm{1}−\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} }\right) \\ $$$$−{v}\frac{{dv}}{{dx}}={g}\left(\mathrm{1}−\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} }\right) \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−{g}\int_{{h}} ^{{x}} \left(\mathrm{1}−\frac{{h}+{h}_{\mathrm{0}} }{{x}+{h}_{\mathrm{0}} }\right){dx} \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−{g}\int_{{h}+{h}_{\mathrm{0}} } ^{{x}+{h}_{\mathrm{0}} } \left(\mathrm{1}−\frac{{h}+{h}_{\mathrm{0}} }{{u}}\right){du} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=−{g}\left[\left({x}−{h}\right)−\left({h}+{h}_{\mathrm{0}} \right)\mathrm{ln}\:\frac{{x}+{h}_{\mathrm{0}} }{{h}+{h}_{\mathrm{0}} }\right] \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{g}\left[\left({h}−{x}\right)+\left({h}+{h}_{\mathrm{0}} \right)\mathrm{ln}\:\frac{{x}+{h}_{\mathrm{0}} }{{h}+{h}_{\mathrm{0}} }\right]} \\ $$$${since}\:{the}\:{time}\:{t}\:{is}\:{infinite},\:{we}\:{can} \\ $$$${not}\:{find}\:{a}\:{and}\:{v}\:{in}\:{terms}\:{of}\:{t}. \\ $$
Commented by fantastic2 last updated on 29/Dec/25
$${great}\:{work} \\ $$