Mechanics Questions

Question Number 100928 by ajfour last updated on 29/Jun/20

Commented by ajfour last updated on 29/Jun/20

$${Find}\:{r}\:{in}\:{terms}\:{of}\:{R}. \\$$

Commented by ajfour last updated on 30/Jun/20

$${Its}\:{physics}\:{plus}\:{maths},\:{Sir}. \\$$

Answered by bramlex last updated on 29/Jun/20

Commented by bramlex last updated on 29/Jun/20

$${x}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \\$$$${x}^{\mathrm{2}} −\mathrm{2}{rR}\:=\:\mathrm{2}{rR}\:\rightarrow{x}\:=\:\mathrm{2}\sqrt{{rR}} \\$$$$\left(\mathrm{1}\right)\left({x}−{R}\right)^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\$$$$\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{Rx}−\mathrm{2}{Ry}+{R}^{\mathrm{2}} =\mathrm{0} \\$$$$\left(\mathrm{2}\right)\left({x}−\left(\mathrm{2}\sqrt{{rR}}+{r}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \right. \\$$$$\rightarrow{x}^{\mathrm{2}} −\mathrm{2}\left({r}+\mathrm{2}\sqrt{{rR}}\right){x}+\left({r}+\mathrm{2}\sqrt{{rR}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ry}=\mathrm{0} \\$$$$... \\$$

Answered by mr W last updated on 29/Jun/20

Commented by mr W last updated on 29/Jun/20

$${at}\:{position}\:\theta: \\$$$$\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} ={R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){mg} \\$$$$\Rightarrow{u}=\sqrt{\mathrm{2}{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){g}} \\$$$${mg}\:\mathrm{cos}\:\theta−{N}=\frac{{mu}^{\mathrm{2}} }{{R}} \\$$$${N}=\mathrm{0}\:{when}\:{contact}\:{gets}\:{lost} \\$$$${mg}\:\mathrm{cos}\:\theta=\frac{{m}×\mathrm{2}{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right){g}}{{R}} \\$$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\$$$$\Rightarrow{u}=\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}} \\$$$${track}\:{of}\:{mass}\:{m}\:{after}\:{point}\:{B}: \\$$$${x}={R}\:\mathrm{sin}\:\theta+{u}\:\mathrm{cos}\:\theta\:{t} \\$$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}{t} \\$$$${y}={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)−{u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\$$$$\Rightarrow{y}=\frac{\mathrm{5}{R}}{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\$$$${D}\left(\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} },\:{r}\right)\equiv{D}\left(\mathrm{2}\sqrt{{Rr}},{r}\right) \\$$$${let}\:\Phi={DC}^{\mathrm{2}} \\$$$$\Phi=\left({x}−\mathrm{2}\sqrt{{Rr}}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} \\$$$$\Phi=\left(\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}{t}−\mathrm{2}\sqrt{{Rr}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}{R}}{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} \\$$$$\frac{{d}\Phi}{{dt}}=\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}{t}−\mathrm{2}\sqrt{{Rr}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}\right)+\mathrm{2}\left(\frac{\mathrm{5}{R}}{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} −{r}\right)\left(−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{Rg}}{\mathrm{3}}}−{gt}\right)=\mathrm{0} \\$$$$\Rightarrow\mathrm{12}\left(\sqrt{\mathrm{30}}+\mathrm{4}\sqrt{\frac{{g}}{{R}}}{t}−\mathrm{6}\sqrt{\mathrm{6}}\sqrt{\frac{{r}}{{R}}}\right)=\left(\mathrm{30}−\mathrm{2}\sqrt{\mathrm{30}}\sqrt{\frac{{g}}{{R}}}\:{t}−\mathrm{9}\left(\sqrt{\frac{{g}}{{R}}}{t}\right)^{\mathrm{2}} −\mathrm{18}\frac{{r}}{{R}}\right)\left(\sqrt{\mathrm{30}}+\mathrm{9}\sqrt{\frac{{g}}{{R}}}{t}\right) \\$$$$\Phi_{{min}} ={r}^{\mathrm{2}} \\$$$$\Rightarrow\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{9}}\sqrt{\frac{{g}}{{R}}}{t}−\mathrm{2}\sqrt{\frac{{r}}{{R}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{3}}−\frac{\sqrt{\mathrm{30}}}{\mathrm{9}}\sqrt{\frac{{g}}{{R}}}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{{g}}{{R}}}{t}\right)^{\mathrm{2}} −\frac{{r}}{{R}}\right)^{\mathrm{2}} =\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} \\$$$${let}\:\delta=\sqrt{\frac{{g}}{{R}}}{t},\:\lambda=\sqrt{\frac{{r}}{{R}}} \\$$$$\Rightarrow\mathrm{12}\left(\sqrt{\mathrm{30}}+\mathrm{4}\delta−\mathrm{6}\sqrt{\mathrm{6}}\lambda\right)=\left(\mathrm{30}−\mathrm{2}\sqrt{\mathrm{30}}\delta−\mathrm{9}\delta^{\mathrm{2}} −\mathrm{18}\lambda^{\mathrm{2}} \right)\left(\sqrt{\mathrm{30}}+\mathrm{9}\delta\right)\:\:\:...\left({i}\right) \\$$$$\Rightarrow\mathrm{4}\left(\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{6}}\delta−\mathrm{18}\lambda\right)^{\mathrm{2}} +\left(\mathrm{30}−\mathrm{2}\sqrt{\mathrm{30}}\delta−\mathrm{9}\delta^{\mathrm{2}} −\mathrm{18}\lambda^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{324}\lambda^{\mathrm{4}} \:\:\:...\left({ii}\right) \\$$$${we}\:{get}\:{from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\$$$$\lambda\approx\mathrm{0}.\mathrm{53716} \\$$$$\Rightarrow{r}=\lambda^{\mathrm{2}} {R}\approx\mathrm{0}.\mathrm{2885}{R} \\$$

Commented by mr W last updated on 29/Jun/20

Commented by ajfour last updated on 29/Jun/20

$${haven}'{t}\:{been}\:{through}\:{all}\:{of}\:{it}\:{sir},\:{but} \\$$$${really}\:{looks}\:{superb}\:{solution},\:{i}\:{want} \\$$$${to}\:{try}\:{on}\:{my}\:{own}\:{for}\:{a}\:{little}\:{while}... \\$$

Commented by ajfour last updated on 29/Jun/20

$${can}\:{you}\:{help}\:{me}\:{sir},\:{with}\:{the}\:{equation} \\$$$${of}\:{parabola}\:{with}\:{shown}\:{axes}..? \\$$

Commented by mr W last updated on 30/Jun/20

Commented by ajfour last updated on 02/Jul/20

$${x}={R}\mathrm{sin}\:\theta+\left({u}\mathrm{cos}\:\theta\right){t} \\$$$${y}={R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)−\left({u}\mathrm{sin}\:\theta\right){t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\$$$${As}\:\:\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}\:,\:\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\:,\:{u}^{\mathrm{2}} =\frac{\mathrm{2}{Rg}}{\mathrm{3}} \\$$$${y}=\frac{\mathrm{5}{R}}{\mathrm{3}}−\left(\frac{{u}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)\frac{\left({x}−\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)}{\left(\frac{\mathrm{2}{u}}{\mathrm{3}}\right)}−\frac{{g}}{\mathrm{2}}\frac{\left({x}−\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\frac{\mathrm{2}{u}}{\mathrm{3}}\right)^{\mathrm{2}} } \\$$$${y}=\frac{\mathrm{5}{R}}{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left({x}−\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)−\frac{\mathrm{27}}{\mathrm{16}{R}}\left({x}−\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)^{\mathrm{2}} \\$$$$\frac{{dy}}{{dx}}=−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{27}}{\mathrm{8}{R}}\left({x}−\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}\right)=\mathrm{0} \\$$$$\Rightarrow\:\:\:{x}_{\mathrm{0}} =\frac{{R}\sqrt{\mathrm{5}}}{\mathrm{3}}−\frac{\mathrm{8}{R}}{\mathrm{27}}\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}} \\$$$${y}_{\mathrm{0}} =\frac{\mathrm{5}{R}}{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left(\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}}−\frac{\mathrm{9}{R}\sqrt{\mathrm{5}}}{\mathrm{27}}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{27}}{\mathrm{16}{R}}\left(\frac{\mathrm{80}{R}^{\mathrm{2}} }{\mathrm{27}×\mathrm{27}}\right) \\$$$$\Rightarrow\:{y}_{\mathrm{0}} =\frac{\mathrm{5}{R}}{\mathrm{3}}+\frac{\mathrm{10}{R}}{\mathrm{27}}−\frac{\mathrm{5}{R}}{\mathrm{27}}=\frac{\mathrm{50}{R}}{\mathrm{27}}\:\:<\:\mathrm{2}{R} \\$$$${Thus}\:{eq}.\:{of}\:{parabola}\:{is} \\$$$$\:\:\boldsymbol{{y}}=\frac{\mathrm{50}\boldsymbol{{R}}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}\boldsymbol{{R}}}\left(\boldsymbol{{x}}−\frac{\mathrm{5}\sqrt{\mathrm{5}}\boldsymbol{{R}}}{\mathrm{27}}\right)^{\mathrm{2}} \\$$$$\:\:\frac{{dy}}{{dx}}=−\frac{\mathrm{27}}{\mathrm{8}{R}}\left({x}−\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}}\right) \\$$$${say}\:{this}\:{parabola}\:{touches}\:{the}\:{smaller} \\$$$${circle}\:{at}\:{P}\:\left({h},{k}\right)\:\:{and}\:{let}\:{center}\:{of} \\$$$${smaller}\:{circle}\:{is}\:{C}\left(\mathrm{2}\sqrt{{Rr}},\:{r}\right);\:\:{then} \\$$$${h}=\mathrm{2}\sqrt{{Rr}}+{r}\mathrm{cos}\:\boldsymbol{\beta} \\$$$${k}={r}+{r}\mathrm{sin}\:\boldsymbol{\beta} \\$$$$\mathrm{cot}\:\beta=\frac{\mathrm{27}}{\mathrm{8}{R}}\left({h}−\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}}\right)=\frac{{h}−\mathrm{2}\sqrt{{Rr}}}{{k}−{r}} \\$$$$\\$$$$\Rightarrow\:\:\:\:\:\frac{\mathrm{27}}{\mathrm{8}}{z}=\frac{\left({z}+\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{27}}\right)−\mathrm{2}\sqrt{{s}}}{\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} −{s}}\:\:\:\:\:\:\:......\left({i}\right) \\$$$$\\$$$$\:\:{k}=\frac{\mathrm{50}{R}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}{R}}\left({h}−\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}}\right)^{\mathrm{2}} \\$$$$\:\:\:\:\:=\:{r}+\frac{{r}}{\sqrt[{}]{\mathrm{1}+\left[\frac{\mathrm{27}}{\mathrm{8}{R}}\left({h}−\frac{\mathrm{5}\sqrt{\mathrm{5}}{R}}{\mathrm{27}}\right)\right]^{\mathrm{2}} }} \\$$$$\\$$$$\Rightarrow\:\:\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} ={s}+\frac{{s}}{\sqrt[{}]{\mathrm{1}+\left(\frac{\mathrm{27}}{\mathrm{16}}{z}\right)^{\mathrm{2}} }}\:\:.....\left({ii}\right) \\$$$$\Rightarrow\:\:{s}=\frac{\left(\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} \right)}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt[{}]{\mathrm{1}+\left(\frac{\mathrm{27}{z}}{\mathrm{16}}\right)^{\mathrm{2}} }}}\:\:\:\:\:\:;\:\:\:{Now} \\$$$$\Rightarrow\:\:\:\:\:\frac{\mathrm{27}}{\mathrm{8}}{z}=\frac{\left({z}+\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{27}}\right)−\mathrm{2}\left\{\frac{\left(\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} \right)}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt[{}]{\mathrm{1}+\left(\frac{\mathrm{27}{z}}{\mathrm{16}}\right)^{\mathrm{2}} }}}\right\}^{\mathrm{1}/\mathrm{2}} }{\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} −\left(\frac{\left(\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} \right)}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt[{}]{\mathrm{1}+\left(\frac{\mathrm{27}{z}}{\mathrm{16}}\right)^{\mathrm{2}} }}}\right)}\:\:\:\:\:\:\:......\left({A}\right) \\$$$${z}\:{is}\:{obtained}\:{from}\:{above}\:{eq}. \\$$$${Then}\:\:\:{s}=\frac{{r}}{{R}}\:=\:\frac{\left(\frac{\mathrm{50}}{\mathrm{27}}−\frac{\mathrm{27}}{\mathrm{16}}{z}^{\mathrm{2}} \right)}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt[{}]{\mathrm{1}+\left(\frac{\mathrm{27}{z}}{\mathrm{16}}\right)^{\mathrm{2}} }}}\:\:\blacksquare \\$$$$\\$$$$\\$$$$\\$$

Commented by ajfour last updated on 02/Jul/20

Commented by mr W last updated on 02/Jul/20

$${beautifully}\:{solved}! \\$$