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Question Number 127811 by mr W last updated on 02/Jan/21

Commented by mr W last updated on 02/Jan/21

$$aparaboloidbowlhasadepthofH$$$$andanopeningwithdiameterL.$$$$asmallballofmassmisreleased$$$$fromrestattherimofthebowl.$$$$assumethesurfaceisfrictionless.$$$$Q1:cantheballreachtherimon$$$$theotherside?$$$$Q2:iftheanswertoQ1isyes,find$$$$thetimetheballneedstoreachthe$$$$rimontheotherside.$$$$Q3:iftheanswertoQ1isno,find$$$$thepositionwheretheballlosses$$$$contactwiththeinnersurfaceofthe$$$$bowl.$$

Answered by mr W last updated on 03/Jan/21

$$whentheballmovesalongtheinner$$$$surfaceofthebowl,thecomponent$$$$ofthegravityforceinnormal$$$$directionatanypointisinthesame$$$$directionasthecentrifugalforce,$$$$bothpresstheballagainsttheinner$$$$wallofthebowl.thatmeansatany$$$$pointthereexistsanormalcontact$$$$forcebetweenthebowlandtheball,$$$$i.e.theballneverlossescontactto$$$$thebowl.$$$$sincethereisnofriction,afterthe$$$$ballisreleasedfromrestattherim$$$$oftheball,itcanreachexactlythe$$$$sameheightasatitsoriginalposition,$$$$thatmeans,theballcanexactlyreach$$$$therimontheoppositeside.$$$$sotheanswertoQ1isyes.$$

Commented by mr W last updated on 03/Jan/21

Commented by mr W last updated on 03/Jan/21

$$letR=\frac{L}{2}=radiusofopening$$$$y=H{\left(\frac{x}{R}\right)}^2$$$$let\eta =\frac{H}{R},\xi =\frac{x}{R}\in [-1,1]$$$$y^{\prime }=2H\frac{x}{R^2}=2\eta \xi$$$$y^{″}=2\frac{H}{R^2}=\frac{2\eta }{R}$$$$radiusofcurvaturer=\frac{{[1+{\left(y^{\prime }\right)}^2]}^{\frac{3}{2}}}{\mid y^{″}\mid }$$$$\Rightarrow r=\frac{{(1+4{\eta }^2\xi )}^{\frac{3}{2}}R}{2\eta }$$$$\frac{1}{2}{mv}^2=mg(H-y)=mg(H-H{\xi }^2)$$$$\Rightarrow v^2=2gH(1-{\xi }^2)$$$$\Rightarrow v=\sqrt{2gH(1-{\xi }^2)}$$$$$$$$ds=\sqrt{1+{\left(y^{\prime }\right)}^2}dx=R\sqrt{1+4{\eta }^2{\xi }^2}d\xi$$$$dt=\frac{ds}{v}=\frac{R}{\sqrt{2gH}}\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$thetimetheballneedstomovefrom$$$$rimtorimisT.$$$$T=\frac{R}{\sqrt{2gH}}{\int }_{-1}^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$=\frac{2R}{\sqrt{2gH}}{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$\Rightarrow T=\sqrt{\frac{2H}{g}}\times \frac{1}{\eta }{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi =k\sqrt{\frac{2H}{g}}$$$$withk=\frac{1}{\eta }{\int }_0^1\sqrt{\frac{1+4{\eta }^2{\xi }^2}{1-{\xi }^2}}d\xi$$$$note:\sqrt{\frac{2H}{g}}isthetimeforfreefall$$$$oftheballfromheightH.{it}^{\prime }sobvious$$$$thatk>2.$$$$$$$$anexactformulaforkisnotpossible.$$$$examples:$$$$\eta =\frac{H}{R}=1\Rightarrow k\approx 2.635184$$$$\eta =\frac{H}{R}=2\Rightarrow k\approx 2.203569$$$$\eta =\frac{H}{R}=4\Rightarrow k\approx 2.061882$$$$\eta =\frac{H}{R}=10\Rightarrow k\approx 2.012202$$$$$$$$\tan \theta =\frac{dy}{dx}=2\eta \xi$$$$N=mg\cos \theta +m\frac{v^2}{r}$$$$\frac{N}{mg}=\frac{1}{\sqrt{1+4{\eta }^2{\xi }^2}}+\frac{4{\eta }^2(1-{\xi }^2)}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}}$$$$=\frac{1}{\sqrt{1+4{\eta }^2{\xi }^2}}+\frac{4{\eta }^2+1-(1+4{\eta }^2{\xi }^2)}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}}$$$$\Rightarrow \frac{N}{mg}=\frac{1+4{\eta }^2}{{(1+4{\eta }^2{\xi }^2)}^{\frac{3}{2}}}>0$$$$at\xi =\pm 1:N_{min}=\frac{mg}{\sqrt{1+4{\eta }^2}}$$$$at\xi =0:N_{max}=(1+4{\eta }^2)mg$$

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