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Question-127811




Question Number 127811 by mr W last updated on 02/Jan/21
Commented by mr W last updated on 02/Jan/21
a paraboloid bowl has a depth of H  and an opening with diameter L.  a small ball of mass m is released  from rest at the rim of the bowl.  assume the surface is frictionless.  Q1: can the ball reach the rim on  the other side?  Q2: if the answer to Q1 is yes, find  the time the ball needs to reach the  rim on the other side.  Q3: if the answer to Q1 is no, find  the position where the ball losses  contact with the inner surface of the  bowl.
$${a}\:{paraboloid}\:{bowl}\:{has}\:{a}\:{depth}\:{of}\:{H} \\ $$$${and}\:{an}\:{opening}\:{with}\:{diameter}\:{L}. \\ $$$${a}\:{small}\:{ball}\:{of}\:{mass}\:{m}\:{is}\:{released} \\ $$$${from}\:{rest}\:{at}\:{the}\:{rim}\:{of}\:{the}\:{bowl}. \\ $$$${assume}\:{the}\:{surface}\:{is}\:{frictionless}. \\ $$$${Q}\mathrm{1}:\:{can}\:{the}\:{ball}\:{reach}\:{the}\:{rim}\:{on} \\ $$$${the}\:{other}\:{side}? \\ $$$${Q}\mathrm{2}:\:{if}\:{the}\:{answer}\:{to}\:{Q}\mathrm{1}\:{is}\:{yes},\:{find} \\ $$$${the}\:{time}\:{the}\:{ball}\:{needs}\:{to}\:{reach}\:{the} \\ $$$${rim}\:{on}\:{the}\:{other}\:{side}. \\ $$$${Q}\mathrm{3}:\:{if}\:{the}\:{answer}\:{to}\:{Q}\mathrm{1}\:{is}\:{no},\:{find} \\ $$$${the}\:{position}\:{where}\:{the}\:{ball}\:{losses} \\ $$$${contact}\:{with}\:{the}\:{inner}\:{surface}\:{of}\:{the} \\ $$$${bowl}. \\ $$
Answered by mr W last updated on 03/Jan/21
when the ball moves along the inner  surface of the bowl, the component  of the gravity force in normal   direction at any point is in the same  direction as the centrifugal force,  both press the ball against the inner  wall of the bowl. that means at any  point there exists a normal contact  force between the bowl and the ball,  i.e. the ball never losses contact to  the bowl.  since there is no friction, after the  ball is released from rest at the rim  of the ball, it can reach exactly the  same height as at its original position,  that means, the ball can exactly reach  the rim on the opposite side.  so the answer to Q1 is yes.
$${when}\:{the}\:{ball}\:{moves}\:{along}\:{the}\:{inner} \\ $$$${surface}\:{of}\:{the}\:{bowl},\:{the}\:{component} \\ $$$${of}\:{the}\:{gravity}\:{force}\:{in}\:{normal}\: \\ $$$${direction}\:{at}\:{any}\:{point}\:{is}\:{in}\:{the}\:{same} \\ $$$${direction}\:{as}\:{the}\:{centrifugal}\:{force}, \\ $$$${both}\:{press}\:{the}\:{ball}\:{against}\:{the}\:{inner} \\ $$$${wall}\:{of}\:{the}\:{bowl}.\:{that}\:{means}\:{at}\:{any} \\ $$$${point}\:{there}\:{exists}\:{a}\:{normal}\:{contact} \\ $$$${force}\:{between}\:{the}\:{bowl}\:{and}\:{the}\:{ball}, \\ $$$${i}.{e}.\:{the}\:{ball}\:{never}\:{losses}\:{contact}\:{to} \\ $$$${the}\:{bowl}. \\ $$$${since}\:{there}\:{is}\:{no}\:{friction},\:{after}\:{the} \\ $$$${ball}\:{is}\:{released}\:{from}\:{rest}\:{at}\:{the}\:{rim} \\ $$$${of}\:{the}\:{ball},\:{it}\:{can}\:{reach}\:{exactly}\:{the} \\ $$$${same}\:{height}\:{as}\:{at}\:{its}\:{original}\:{position}, \\ $$$${that}\:{means},\:{the}\:{ball}\:{can}\:{exactly}\:{reach} \\ $$$${the}\:{rim}\:{on}\:{the}\:{opposite}\:{side}. \\ $$$${so}\:{the}\:{answer}\:{to}\:{Q}\mathrm{1}\:{is}\:{yes}. \\ $$
Commented by mr W last updated on 03/Jan/21
Commented by mr W last updated on 03/Jan/21
let R=(L/2)=radius of opening  y=H((x/R))^2   let η=(H/R), ξ=(x/R)∈[−1,1]  y′=2H(x/R^2 )=2ηξ  y′′=2(H/R^2 )=((2η)/R)  radius of curvature r=(([1+(y′)^2 ]^(3/2) )/(∣y′′∣))  ⇒r=(((1+4η^2 ξ)^(3/2) R)/(2η))  (1/2)mv^2 =mg(H−y)=mg(H−Hξ^2 )  ⇒v^2 =2gH(1−ξ^2 )  ⇒v=(√(2gH(1−ξ^2 )))    ds=(√(1+(y′)^2 ))dx=R(√(1+4η^2 ξ^2 ))dξ  dt=(ds/v)=(R/( (√(2gH))))(√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ  the time the ball needs to move from  rim to rim is T.  T=(R/( (√(2gH))))∫_(−1) ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ  =((2R)/( (√(2gH))))∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ  ⇒T=(√((2H)/g))×(1/η)∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ=k(√((2H)/g))  with k=(1/η)∫_0 ^1 (√((1+4η^2 ξ^2 )/(1−ξ^2 )))dξ  note: (√((2H)/g)) is the time for free fall  of the ball from height H. it′s obvious  that k>2.    an exact formula for k is not possible.  examples:  η=(H/R)=1 ⇒k≈2.635184  η=(H/R)=2 ⇒k≈2.203569  η=(H/R)=4 ⇒k≈2.061882  η=(H/R)=10 ⇒k≈2.012202    tan θ=(dy/dx)=2ηξ  N=mg cos θ+m(v^2 /r)  (N/(mg))=(1/( (√(1+4η^2 ξ^2 ))))+((4η^2 (1−ξ^2 ))/((1+4η^2 ξ^2 )^(3/2) ))  =(1/( (√(1+4η^2 ξ^2 ))))+((4η^2 +1−(1+4η^2 ξ^2 ))/((1+4η^2 ξ^2 )^(3/2) ))  ⇒(N/(mg))=((1+4η^2 )/((1+4η^2 ξ^2 )^(3/2) )) >0  at ξ=±1: N_(min) =((mg)/( (√(1+4η^2 ))))  at ξ=0: N_(max) =(1+4η^2 )mg
$${let}\:{R}=\frac{{L}}{\mathrm{2}}={radius}\:{of}\:{opening} \\ $$$${y}={H}\left(\frac{{x}}{{R}}\right)^{\mathrm{2}} \\ $$$${let}\:\eta=\frac{{H}}{{R}},\:\xi=\frac{{x}}{{R}}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${y}'=\mathrm{2}{H}\frac{{x}}{{R}^{\mathrm{2}} }=\mathrm{2}\eta\xi \\ $$$${y}''=\mathrm{2}\frac{{H}}{{R}^{\mathrm{2}} }=\frac{\mathrm{2}\eta}{{R}} \\ $$$${radius}\:{of}\:{curvature}\:{r}=\frac{\left[\mathrm{1}+\left({y}'\right)^{\mathrm{2}} \right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mid{y}''\mid} \\ $$$$\Rightarrow{r}=\frac{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {R}}{\mathrm{2}\eta} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mg}\left({H}−{y}\right)={mg}\left({H}−{H}\xi^{\mathrm{2}} \right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\mathrm{2}{gH}\left(\mathrm{1}−\xi^{\mathrm{2}} \right) \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{gH}\left(\mathrm{1}−\xi^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${ds}=\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}={R}\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{d}\xi \\ $$$${dt}=\frac{{ds}}{{v}}=\frac{{R}}{\:\sqrt{\mathrm{2}{gH}}}\sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$${the}\:{time}\:{the}\:{ball}\:{needs}\:{to}\:{move}\:{from} \\ $$$${rim}\:{to}\:{rim}\:{is}\:{T}. \\ $$$${T}=\frac{{R}}{\:\sqrt{\mathrm{2}{gH}}}\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$$=\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{2}{gH}}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$$\Rightarrow{T}=\sqrt{\frac{\mathrm{2}{H}}{{g}}}×\frac{\mathrm{1}}{\eta}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi={k}\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$${with}\:{k}=\frac{\mathrm{1}}{\eta}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} }}{d}\xi \\ $$$${note}:\:\sqrt{\frac{\mathrm{2}{H}}{{g}}}\:{is}\:{the}\:{time}\:{for}\:{free}\:{fall} \\ $$$${of}\:{the}\:{ball}\:{from}\:{height}\:{H}.\:{it}'{s}\:{obvious} \\ $$$${that}\:{k}>\mathrm{2}. \\ $$$$ \\ $$$${an}\:{exact}\:{formula}\:{for}\:{k}\:{is}\:{not}\:{possible}. \\ $$$${examples}: \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{1}\:\Rightarrow{k}\approx\mathrm{2}.\mathrm{635184} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{2}\:\Rightarrow{k}\approx\mathrm{2}.\mathrm{203569} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{4}\:\Rightarrow{k}\approx\mathrm{2}.\mathrm{061882} \\ $$$$\eta=\frac{{H}}{{R}}=\mathrm{10}\:\Rightarrow{k}\approx\mathrm{2}.\mathrm{012202} \\ $$$$ \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\mathrm{2}\eta\xi \\ $$$${N}={mg}\:\mathrm{cos}\:\theta+{m}\frac{{v}^{\mathrm{2}} }{{r}} \\ $$$$\frac{{N}}{{mg}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }}+\frac{\mathrm{4}\eta^{\mathrm{2}} \left(\mathrm{1}−\xi^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} }}+\frac{\mathrm{4}\eta^{\mathrm{2}} +\mathrm{1}−\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \xi^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:>\mathrm{0} \\ $$$${at}\:\xi=\pm\mathrm{1}:\:{N}_{{min}} =\frac{{mg}}{\:\sqrt{\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} }} \\ $$$${at}\:\xi=\mathrm{0}:\:{N}_{{max}} =\left(\mathrm{1}+\mathrm{4}\eta^{\mathrm{2}} \right){mg} \\ $$

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