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Question-126080




Question Number 126080 by ajfour last updated on 17/Dec/20
Commented by ajfour last updated on 18/Dec/20
Cylinder and sphere have the  same radius. An equilateral  triangular plate rests on sphere  with two vertices against curved  surface of cylinder and top vertex  on rim of cylinder. It is in contact  with sphere at point P which is at  a height h from base of cylinder.  (PQ=h).  Find height of cylinder.
$${Cylinder}\:{and}\:{sphere}\:{have}\:{the} \\ $$$${same}\:{radius}.\:{An}\:{equilateral} \\ $$$${triangular}\:{plate}\:{rests}\:{on}\:{sphere} \\ $$$${with}\:{two}\:{vertices}\:{against}\:{curved} \\ $$$${surface}\:{of}\:{cylinder}\:{and}\:{top}\:{vertex} \\ $$$${on}\:{rim}\:{of}\:{cylinder}.\:{It}\:{is}\:{in}\:{contact} \\ $$$${with}\:{sphere}\:{at}\:{point}\:{P}\:{which}\:{is}\:{at} \\ $$$${a}\:{height}\:\boldsymbol{{h}}\:{from}\:{base}\:{of}\:{cylinder}. \\ $$$$\left({PQ}={h}\right).\:\:{Find}\:{height}\:{of}\:{cylinder}. \\ $$
Commented by mr W last updated on 17/Dec/20
if the top vertex has contact force  with the rim of cylinder, no unique  solution exists. if we assume that  the top vertex is just on the rim, but  without real contact, then i think  there is an unique solution. please  check sir.
$${if}\:{the}\:{top}\:{vertex}\:{has}\:{contact}\:{force} \\ $$$${with}\:{the}\:{rim}\:{of}\:{cylinder},\:{no}\:{unique} \\ $$$${solution}\:{exists}.\:{if}\:{we}\:{assume}\:{that} \\ $$$${the}\:{top}\:{vertex}\:{is}\:{just}\:{on}\:{the}\:{rim},\:{but} \\ $$$${without}\:{real}\:{contact},\:{then}\:{i}\:{think} \\ $$$${there}\:{is}\:{an}\:{unique}\:{solution}.\:{please} \\ $$$${check}\:{sir}. \\ $$
Commented by ajfour last updated on 18/Dec/20
triangle is oriented symmetrically  left and right, and top vertex just  reaching the rim, no contact force.  (I get your point, Sir, lets assume  frictionless surfaces of sphere,  triangular plate and cylinder).
$${triangle}\:{is}\:{oriented}\:{symmetrically} \\ $$$${left}\:{and}\:{right},\:{and}\:{top}\:{vertex}\:{just} \\ $$$${reaching}\:{the}\:{rim},\:{no}\:{contact}\:{force}. \\ $$$$\left({I}\:{get}\:{your}\:{point},\:{Sir},\:{lets}\:{assume}\right. \\ $$$${frictionless}\:{surfaces}\:{of}\:{sphere}, \\ $$$$\left.{triangular}\:{plate}\:{and}\:{cylinder}\right). \\ $$
Answered by mr W last updated on 17/Dec/20
Commented by mr W last updated on 17/Dec/20
s=side length of equilateral triangle  B′=midpoint of BC  D=center of mass of plate  AB′=(((√3)s)/2)  AD=(2/3)×(((√3)s)/2)=(((√3)s)/3)  DB′=(1/3)×(((√3)s)/2)=(((√3)s)/6)  h=r+r sin θ  ⇒sin θ=(h/r)−1  2r=(((√3)s)/2)×sin θ+r−(√(r^2 −(s^2 /4)))  ⇒(((√3)s)/2)×((h/r)−1)−r=(√(r^2 −(s^2 /4)))  ⇒[3((h/r)−1)^2 +1]s=4(√3)((h/r)−1)r  ⇒s=((4(√3)((h/r)−1)r)/(3((h/r)−1)^2 +1))  PD=SD×cos θ=DB′×cos^2  θ  =(((√3)s)/6)[1−((h/r)−1)^2 ]  AP=AD+PD=(((√3)s)/3)+(((√3)s)/6)[1−((h/r)−1)^2 ]  =(((√3)s)/6)[3−((h/r)−1)^2 ]  height of cylinder=H  H=h+AP×cos θ  ⇒H=h+(((√3)s)/6)[3−((h/r)−1)^2 ](√(1−((h/r)−1)^2 ))
$${s}={side}\:{length}\:{of}\:{equilateral}\:{triangle} \\ $$$${B}'={midpoint}\:{of}\:{BC} \\ $$$${D}={center}\:{of}\:{mass}\:{of}\:{plate} \\ $$$${AB}'=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}} \\ $$$${AD}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{3}} \\ $$$${DB}'=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}} \\ $$$${h}={r}+{r}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{h}}{{r}}−\mathrm{1} \\ $$$$\mathrm{2}{r}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}×\mathrm{sin}\:\theta+{r}−\sqrt{{r}^{\mathrm{2}} −\frac{{s}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}×\left(\frac{{h}}{{r}}−\mathrm{1}\right)−{r}=\sqrt{{r}^{\mathrm{2}} −\frac{{s}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\Rightarrow\left[\mathrm{3}\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right]{s}=\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{{h}}{{r}}−\mathrm{1}\right){r} \\ $$$$\Rightarrow{s}=\frac{\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{{h}}{{r}}−\mathrm{1}\right){r}}{\mathrm{3}\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${PD}={SD}×\mathrm{cos}\:\theta={DB}'×\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}}\left[\mathrm{1}−\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$${AP}={AD}+{PD}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}}\left[\mathrm{1}−\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$$=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}}\left[\mathrm{3}−\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$${height}\:{of}\:{cylinder}={H} \\ $$$${H}={h}+{AP}×\mathrm{cos}\:\theta \\ $$$$\Rightarrow{H}={h}+\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{6}}\left[\mathrm{3}−\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} \right]\sqrt{\mathrm{1}−\left(\frac{{h}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 18/Dec/20
Very smart (intelligent) diagram  and solution Sir, loved following  it; TAL (thanks a lot)!
$${Very}\:{smart}\:\left({intelligent}\right)\:{diagram} \\ $$$${and}\:{solution}\:{Sir},\:{loved}\:{following} \\ $$$${it};\:\mathcal{TAL}\:\left({thanks}\:{a}\:{lot}\right)! \\ $$
Commented by mr W last updated on 18/Dec/20
i think in 3D, but can only illustrate  with 2D−diagrams. i like your 3D−  diagrams!
$${i}\:{think}\:{in}\:\mathrm{3}{D},\:{but}\:{can}\:{only}\:{illustrate} \\ $$$${with}\:\mathrm{2}{D}−{diagrams}.\:{i}\:{like}\:{your}\:\mathrm{3}{D}− \\ $$$${diagrams}! \\ $$

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