Question Number 214398 by mr W last updated on 07/Dec/24
Commented by mr W last updated on 07/Dec/24
$${find}\:{acceleration}\:{of}\:{wedge}\:{A}=? \\ $$$${cylinder}\:{with}\:{mass}\:{m}\:{and}\:{radius}\:{r}\: \\ $$$${rolls}\:{along}\:{the}\:{wedge}\:{with}\:{mass}\:{M}. \\ $$
Answered by mr W last updated on 07/Dec/24
Commented by mr W last updated on 08/Dec/24
$${a}={relative}\:{acceleration}\:{of}\:{m} \\ $$$${MA}={N}\:\mathrm{sin}\:\theta+{f}\:\mathrm{cos}\:\theta\:\:\:…\left({i}\right) \\ $$$$−{mA}\:\mathrm{sin}\:\theta={N}−{mg}\:\mathrm{cos}\:\theta\:\:\:…\left({ii}\right) \\ $$$${m}\left({a}−{A}\:\mathrm{cos}\:\theta\right)={mg}\:\mathrm{sin}\:\theta+{f}\:\:\:…\left({iii}\right) \\ $$$$\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}×\frac{{a}}{{r}}=−{fr}\:\:\:…\left({iv}\right) \\ $$$$\Rightarrow{f}=−\frac{{ma}}{\mathrm{2}} \\ $$$${ma}−{mA}\:\mathrm{cos}\:\theta={mg}\:\mathrm{sin}\:\theta−\frac{{ma}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}\left({g}\:\mathrm{sin}\:\theta+{A}\:\mathrm{cos}\:\theta\right)}{\mathrm{3}} \\ $$$$\Rightarrow{N}={m}\left({g}\:\mathrm{cos}\:\theta−{A}\:\mathrm{sin}\:\theta\right) \\ $$$${MA}={m}\left({g}\:\mathrm{cos}\:\theta−{A}\:\mathrm{sin}\:\theta\right)\:\mathrm{sin}\:\theta−\frac{{m}}{\mathrm{2}}×\frac{\mathrm{2}\left({g}\:\mathrm{sin}\:\theta+{A}\:\mathrm{cos}\:\theta\right)}{\mathrm{3}}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{3}\frac{{M}}{{m}}{A}=\mathrm{2}{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta−\mathrm{3}{A}+\mathrm{2}{A}\:\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$\Rightarrow{A}=\frac{{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{M}}{{m}}+\mathrm{1}\right)−\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$