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Question Number 226293 by ajfour last updated on 25/Nov/25

A hemispherical bowl of radius R with maimum water in it without needing to spill is spinning with the content at constant ω. Find volume of water in bowl.

$${A}\:{hemispherical}\:{bowl}\:{of}\:{radius}\:{R} \\ $$$$\:{with}\:{maimum}\:{water}\:{in}\:{it}\:{without} \\ $$$${needing}\:{to}\:{spill}\:{is}\:{spinning}\:{with}\:{the} \\ $$$${content}\:{at}\:{constant}\:\omega.\:{Find}\:{volume} \\ $$$${of}\:{water}\:{in}\:{bowl}. \\ $$$$ \\ $$

Commented by fantastic2 last updated on 25/Nov/25

additional quest1: while spinning , find how much the bottom point of the water will go down from its initial level

$${additional}\:{quest}\mathrm{1}: \\ $$$${while}\:{spinning}\:, \\ $$$${find}\:{how}\:{much} \\ $$$$\:{the}\:{bottom}\:{point} \\ $$$${of}\:{the}\:{water}\:{will}\:{go}\:{down} \\ $$$${from}\:{its}\:{initial}\:{level} \\ $$

Answered by mr W last updated on 25/Nov/25

Commented by mr W last updated on 25/Nov/25

Commented by mr W last updated on 25/Nov/25

mω^2 x=mg tan θ=mgy′ y′=((ω^2 x)/g) ⇒y=((ω^2 x^2 )/(2g)) h=((ω^2 R^2 )/(2g)) (=((4ξR)/3)) V_w =((2πR^3 )/3)−(πR^2 h−2π×((3R)/4)×((Rh)/3)) =((2πR^3 )/3)−((πR^2 h)/2) =((2πR^3 )/3)−((πω^2 R^4 )/(4g)) say initial water depth is h_1 . πh_1 ^2 (R−(h_1 /3))=((2πR^3 )/3)−((πω^2 R^4 )/(4g)) let λ=(h_1 /R), μ=1−((3ω^2 R)/(8g))=1−ξ λ^3 −3λ^2 +2μ=0 λ=s+1 ξ=((3ω^2 R)/(8g)) s^3 −3s−2ξ=0 s=2 sin (−((sin^(−1) ξ)/3)) λ=(h_1 /R)=1−2 sin ((1/3) sin^(−1) ξ) h=h_1 −h_2 =h_1 −(R−h)=h_1 +h−R (( h)/R)=((ω^2 R)/(2g))−2 sin (((sin^(−1) ξ)/3) ) ⇒(( h)/R)=((4ξ)/3)−2 sin (((sin^(−1) ξ)/3) )

$${m}\omega^{\mathrm{2}} {x}={mg}\:\mathrm{tan}\:\theta={mgy}' \\ $$$${y}'=\frac{\omega^{\mathrm{2}} {x}}{{g}}\: \\ $$$$\Rightarrow{y}=\frac{\omega^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{g}} \\ $$$${h}=\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}}\:\:\:\:\left(=\frac{\mathrm{4}\xi{R}}{\mathrm{3}}\right) \\ $$$${V}_{{w}} =\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\left(\pi{R}^{\mathrm{2}} {h}−\mathrm{2}\pi×\frac{\mathrm{3}{R}}{\mathrm{4}}×\frac{{Rh}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi{R}^{\mathrm{2}} {h}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi\omega^{\mathrm{2}} {R}^{\mathrm{4}} }{\mathrm{4}{g}} \\ $$$${say}\:{initial}\:{water}\:{depth}\:{is}\:{h}_{\mathrm{1}} . \\ $$$$\:\pi{h}_{\mathrm{1}} ^{\mathrm{2}} \left({R}−\frac{{h}_{\mathrm{1}} }{\mathrm{3}}\right)=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi\omega^{\mathrm{2}} {R}^{\mathrm{4}} }{\mathrm{4}{g}} \\ $$$${let}\:\lambda=\frac{{h}_{\mathrm{1}} }{{R}},\:\mu=\mathrm{1}−\frac{\mathrm{3}\omega^{\mathrm{2}} {R}}{\mathrm{8}{g}}=\mathrm{1}−\xi \\ $$$$\:\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\mu=\mathrm{0} \\ $$$$\lambda={s}+\mathrm{1} \\ $$$$\xi=\frac{\mathrm{3}\omega^{\mathrm{2}} {R}}{\mathrm{8}{g}} \\ $$$${s}^{\mathrm{3}} −\mathrm{3}{s}−\mathrm{2}\xi=\mathrm{0} \\ $$$${s}=\mathrm{2}\:\mathrm{sin}\:\left(−\frac{\mathrm{sin}^{−\mathrm{1}} \xi}{\mathrm{3}}\right)\: \\ $$$$\lambda=\frac{{h}_{\mathrm{1}} }{{R}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \xi\right) \\ $$$$\:{h}={h}_{\mathrm{1}} −{h}_{\mathrm{2}} ={h}_{\mathrm{1}} −\left({R}−{h}\right)={h}_{\mathrm{1}} +{h}−{R} \\ $$$$\frac{\:{h}}{{R}}=\frac{\omega^{\mathrm{2}} {R}}{\mathrm{2}{g}}−\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{sin}^{−\mathrm{1}} \xi}{\mathrm{3}}\:\right) \\ $$$$\Rightarrow\frac{\:{h}}{{R}}=\frac{\mathrm{4}\xi}{\mathrm{3}}−\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{sin}^{−\mathrm{1}} \xi}{\mathrm{3}}\:\right) \\ $$

Commented by fantastic2 last updated on 25/Nov/25

sir pls explain 5th line

$${sir}\:{pls}\:{explain}\:\mathrm{5}{th}\:{line} \\ $$

Commented by fantastic2 last updated on 25/Nov/25

oo i understand it now sir

$${oo}\:{i}\:{understand}\:{it}\:{now}\:{sir} \\ $$

Commented by mr W last updated on 25/Nov/25

Commented by fantastic2 last updated on 25/Nov/25

��

Commented by mr W last updated on 25/Nov/25

Commented by ajfour last updated on 25/Nov/25

Thank you sir. You have explained awesome.

$${Thank}\:{you}\:{sir}.\:{You}\:{have}\:{explained} \\ $$$${awesome}. \\ $$

Commented by fantastic2 last updated on 26/Nov/25

$$ \\ $$

Commented by mr W last updated on 26/Nov/25

(what the second diagram shows) minimum volume of water needed to cover the inside of bowl with water through rotating: h=((ω^2 R^2 )/(2g))=R ⇒ω=(√((2g)/R)) V_(min) =((2πR^3 )/3)−((πω^2 R^4 )/(4g))=((2πR^3 )/3)−((πR^3 )/2) =((πR^3 )/6) corresponding minimum depth of water: h_(min) =[1−2 sin ((1/3) sin^(−1) (3/4))]R ≈0.442 R

$$\left({what}\:{the}\:{second}\:{diagram}\:{shows}\right) \\ $$$${minimum}\:{volume}\:{of}\:{water}\:{needed} \\ $$$${to}\:{cover}\:{the}\:{inside}\:{of}\:{bowl}\:{with}\: \\ $$$${water}\:{through}\:{rotating}: \\ $$$${h}=\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}}={R}\:\:\Rightarrow\omega=\sqrt{\frac{\mathrm{2}{g}}{{R}}} \\ $$$$\:{V}_{{min}} =\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi\omega^{\mathrm{2}} {R}^{\mathrm{4}} }{\mathrm{4}{g}}=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi{R}^{\mathrm{3}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\pi{R}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${corresponding}\:{minimum}\:{depth}\: \\ $$$${of}\:{water}: \\ $$$${h}_{{min}} =\left[\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\right)\right]{R} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{442}\:{R} \\ $$

Answered by fantastic2 last updated on 25/Nov/25

Commented by fantastic2 last updated on 25/Nov/25

Ncos θ=mg Nsin θ=mω^2 x ⇒tan θ=((ω^2 x)/g) ⇒(dy/dx)=((ω^2 x)/g)⇒∫dy=∫(ω^2 /g)xdx⇒y=((ω^2 x^2 )/(2g))[C=0] at y=z,x^2 =((2gz)/ω^2 ) A=πx^2 =π((2gz)/ω^2 ) at x=R ,y=((ω^2 R^2 )/(2g)) V_(blank) =∫_0 ^((ω^2 R^2 )/(2g)) π((2gz)/ω^2 )dz=((2gπ)/ω^2 )[(z^2 /2)]_0 ^((ω^2 R^2 )/(2g)) V_(blank) =((ω^2 R^4 )/(2g))π V_(water) +V_(blank) =(2/3)πR^3 V_(water) =(2/3)πR^3 −((ω^2 R^4 )/(2g))π Additional quest 1 i again used integration lets assume the initially height h is filled from the bottom V=∫_0 ^h π(R^2 −(R−z)^2 )dz V=πRh^2 −(π/3)h^3 h^3 −3Rh^2 +((3V)/π)=0 ⇒solve for h∈R Δh=((ω^2 R^2 )/(2g))−(R−h)

$${N}\mathrm{cos}\:\theta={mg} \\ $$$${N}\mathrm{sin}\:\theta={m}\omega^{\mathrm{2}} {x} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\omega^{\mathrm{2}} {x}}{{g}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\omega^{\mathrm{2}} {x}}{{g}}\Rightarrow\int{dy}=\int\frac{\omega^{\mathrm{2}} }{{g}}{xdx}\Rightarrow{y}=\frac{\omega^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{g}}\left[{C}=\mathrm{0}\right] \\ $$$${at}\:{y}={z},{x}^{\mathrm{2}} =\frac{\mathrm{2}{gz}}{\omega^{\mathrm{2}} } \\ $$$${A}=\pi{x}^{\mathrm{2}} =\pi\frac{\mathrm{2}{gz}}{\omega^{\mathrm{2}} } \\ $$$${at}\:{x}={R}\:,{y}=\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}} \\ $$$${V}_{{blank}} =\int_{\mathrm{0}} ^{\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}}} \pi\frac{\mathrm{2}{gz}}{\omega^{\mathrm{2}} }{dz}=\frac{\mathrm{2}{g}\pi}{\omega^{\mathrm{2}} }\left[\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}}} \\ $$$${V}_{{blank}} =\frac{\omega^{\mathrm{2}} {R}^{\mathrm{4}} }{\mathrm{2}{g}}\pi \\ $$$${V}_{{water}} +{V}_{{blank}} =\frac{\mathrm{2}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \\ $$$${V}_{{water}} =\frac{\mathrm{2}}{\mathrm{3}}\pi{R}^{\mathrm{3}} −\frac{\omega^{\mathrm{2}} {R}^{\mathrm{4}} }{\mathrm{2}{g}}\pi \\ $$$${Additional}\:{quest}\:\mathrm{1} \\ $$$${i}\:{again}\:{used}\:{integration} \\ $$$${lets}\:{assume}\:{the}\:{initially} \\ $$$${height}\:{h}\:{is}\:{filled}\:{from}\:{the}\:{bottom} \\ $$$${V}=\int_{\mathrm{0}} ^{{h}} \pi\left({R}^{\mathrm{2}} −\left({R}−{z}\right)^{\mathrm{2}} \right){dz} \\ $$$${V}=\pi{Rh}^{\mathrm{2}} −\frac{\pi}{\mathrm{3}}{h}^{\mathrm{3}} \\ $$$${h}^{\mathrm{3}} −\mathrm{3}{Rh}^{\mathrm{2}} +\frac{\mathrm{3}{V}}{\pi}=\mathrm{0} \\ $$$$\Rightarrow{solve}\:{for}\:{h}\in\mathbb{R} \\ $$$$\Delta{h}=\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{2}{g}}−\left({R}−{h}\right) \\ $$

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