Question Number 53593 by ajfour last updated on 23/Jan/19
Commented by ajfour last updated on 23/Jan/19
![Having removed the largest cube from a hemisphere, it can rest in equilibrium on a rough incline plane of inclination α, in the manner shown. Find θ. (source: Statics & Dynamics by S.L. Loney) Answer given: θ=sin^(−1) [((8(3π−(√6)))/(9π−8))sin α].](Q53594.png)
$${Having}\:{removed}\:{the}\:{largest}\:{cube} \\ $$$${from}\:{a}\:{hemisphere},\:{it}\:{can}\:{rest}\:{in} \\ $$$${equilibrium}\:{on}\:{a}\:{rough}\:{incline} \\ $$$${plane}\:{of}\:{inclination}\:\alpha,\:{in}\:{the} \\ $$$${manner}\:{shown}.\:{Find}\:\theta. \\ $$$$\:\left({source}:\:{Statics}\:\&\:{Dynamics}\right. \\ $$$$\left.\:\:\:\:\:{by}\:{S}.{L}.\:{Loney}\right) \\ $$$${Answer}\:{given}:\:\:\theta=\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mathrm{8}\left(\mathrm{3}\pi−\sqrt{\mathrm{6}}\right)}{\mathrm{9}\pi−\mathrm{8}}\mathrm{sin}\:\alpha\right]. \\ $$
Answered by mr W last updated on 23/Jan/19
Commented by mr W last updated on 23/Jan/19
![R=radius of sphere a^2 +a^2 +(2a)^2 =R^2 ⇒a=(R/(√6)) M=mass of hemisphere m=mass of cube M=((2πR^3 ρ)/3) m=8a^3 ρ=((4R^3 ρ)/(3(√6)))=((2M)/(π(√6))) center of mass of hemisphere: b=((3R)/8) center of mass of hemisphere with hole: c=((Mb−ma)/(M−m))=((((M3R)/8)−((2M)/(π(√6)))×(R/(√6)))/(M−((2M)/(π(√6)))))=((R(9π−8))/(8(3π−(√6)))) (c/(sin α))=(R/(sin (π−θ))) ⇒sin θ=(R/c) sin α=((8(3π−(√6)))/(9π−8))×sin α ⇒θ=sin^(−1) [((8(3π−(√6)))/(9π−8))×sin α]](Q53598.png)
$${R}={radius}\:{of}\:{sphere} \\ $$$${a}^{\mathrm{2}} +{a}^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{R}}{\sqrt{\mathrm{6}}} \\ $$$${M}={mass}\:{of}\:{hemisphere} \\ $$$${m}={mass}\:{of}\:{cube} \\ $$$${M}=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$$${m}=\mathrm{8}{a}^{\mathrm{3}} \rho=\frac{\mathrm{4}{R}^{\mathrm{3}} \rho}{\mathrm{3}\sqrt{\mathrm{6}}}=\frac{\mathrm{2}{M}}{\pi\sqrt{\mathrm{6}}} \\ $$$${center}\:{of}\:{mass}\:{of}\:{hemisphere}: \\ $$$${b}=\frac{\mathrm{3}{R}}{\mathrm{8}} \\ $$$$ \\ $$$${center}\:{of}\:{mass}\:{of}\:{hemisphere}\:{with}\:{hole}: \\ $$$${c}=\frac{{Mb}−{ma}}{{M}−{m}}=\frac{\frac{{M}\mathrm{3}{R}}{\mathrm{8}}−\frac{\mathrm{2}{M}}{\pi\sqrt{\mathrm{6}}}×\frac{{R}}{\sqrt{\mathrm{6}}}}{{M}−\frac{\mathrm{2}{M}}{\pi\sqrt{\mathrm{6}}}}=\frac{{R}\left(\mathrm{9}\pi−\mathrm{8}\right)}{\mathrm{8}\left(\mathrm{3}\pi−\sqrt{\mathrm{6}}\right)} \\ $$$$\frac{{c}}{\mathrm{sin}\:\alpha}=\frac{{R}}{\mathrm{sin}\:\left(\pi−\theta\right)} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{R}}{{c}}\:\mathrm{sin}\:\alpha=\frac{\mathrm{8}\left(\mathrm{3}\pi−\sqrt{\mathrm{6}}\right)}{\mathrm{9}\pi−\mathrm{8}}×\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \left[\frac{\mathrm{8}\left(\mathrm{3}\pi−\sqrt{\mathrm{6}}\right)}{\mathrm{9}\pi−\mathrm{8}}×\mathrm{sin}\:\alpha\right] \\ $$
Commented by ajfour last updated on 23/Jan/19
$${Wonderful}\:{question}\:{and}\:{your}\:{solution}, \\ $$$${Sir}\:! \\ $$
Commented by mr W last updated on 23/Jan/19
$${thank}\:{you}\:{sir}! \\ $$$${in}\:{my}\:{solution}\:{the}\:{side}\:{length}\:{of}\:{cube} \\ $$$${is}\:\mathrm{2}{a},\:{therefore}\:{i}\:{got}\:{a}=\frac{{R}}{\sqrt{\mathrm{6}}}. \\ $$
Commented by ajfour last updated on 23/Jan/19
$$\left.{thanks}\:_{\mid} \underset{\underset{\smile} {-}} {\overset{\overset{\frown} {\frown}} {.\:._{\:\mid} }}\right)\:{Sir}. \\ $$
Answered by ajfour last updated on 23/Jan/19
Commented by mr W last updated on 23/Jan/19
$${nice}\:{diagram}! \\ $$$${i}\:{didn}'{t}\:{make}\:{such}\:{a}\:{diagram},\:{so}\:{i} \\ $$$${made}\:{a}\:{mistake}\:{at}\:{first}. \\ $$