Question Number 39776 by ajfour last updated on 10/Jul/18
Commented by ajfour last updated on 11/Jul/18
$${Find}\:{the}\:{acceleration}\:{of}\:{wedge}, \\ $$$${if}\:{ground}\:{is}\:{smooth}\:{while}\:{the} \\ $$$${discs}\:{purely}\:{rolls}\:{on}\:{the}\:{wedge}. \\ $$$${mass}\:{of}\:{wedge}\:{is}\:{M}\:'\:={M}+{m}\:. \\ $$$${The}\:{connecting}\:{rod}\:{is}\:{light}\:{and} \\ $$$${of}\:{length}\:{l}. \\ $$
Commented by ajfour last updated on 11/Jul/18
Commented by ajfour last updated on 11/Jul/18
$${MgR}\mathrm{sin}\:\alpha+{MAR}\mathrm{cos}\:\alpha−{FR}\mathrm{cos}\:\delta\:=\:\frac{\mathrm{3}}{\mathrm{2}}{MR}^{\mathrm{2}} \left(\frac{{a}}{{R}}\right) \\ $$$${mgr}\mathrm{sin}\:\alpha+{mAr}\mathrm{cos}\:\alpha+{Fr}\mathrm{cos}\:\delta\:=\:\frac{\mathrm{3}}{\mathrm{2}}{mr}^{\mathrm{2}} \left(\frac{{a}}{{r}}\right) \\ $$$$\Rightarrow\:\left({M}+{m}\right){rR}\left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha\right)=\frac{\mathrm{3}{a}}{\mathrm{2}}\left({M}+{m}\right){rR} \\ $$$$\Rightarrow\:\:\boldsymbol{{a}}\:=\:\frac{\mathrm{2}\left(\boldsymbol{{g}}\mathrm{sin}\:\boldsymbol{\alpha}+\boldsymbol{{A}}\mathrm{cos}\:\boldsymbol{\alpha}\right)}{\mathrm{3}} \\ $$$$\mathrm{2}\left({M}+{m}\right){A}=\left({M}+{m}\right){a}\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:{a}=\:\mathrm{2}{A}/\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:\mathrm{6}{A}\:=\:\mathrm{2}{g}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha+\mathrm{2}{A}\mathrm{cos}\:^{\mathrm{2}} \alpha \\ $$$${or}\:\:\:\:\:\:\boldsymbol{{A}}\:=\:\frac{\boldsymbol{{g}}\mathrm{sin}\:\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\alpha}}{\mathrm{3}−\mathrm{2cos}\:^{\mathrm{2}} \boldsymbol{\alpha}}\:. \\ $$