Question Number 110993 by ajfour last updated on 01/Sep/20
Commented by ajfour last updated on 01/Sep/20
$${Assume}\:{all}\:{surfaces}\:{frictionless}. \\ $$$${Find}\:{acceleration}\:{of}\:{center}\:{of}\:{mass} \\ $$$${of}\:{the}\:{system}. \\ $$
Answered by mr W last updated on 01/Sep/20
Commented by mr W last updated on 01/Sep/20
![N sin θ=MA N+mA sin θ=mg cos θ mg sin θ+mA cos θ=ma ⇒g sin θ+A cos θ=a A((M/m)+sin^2 θ)=g sin θ cos θ ⇒A=((g sin θ cos θ)/((M/m)+sin^2 θ))=(dx_M ^2 /dt^2 ) ⇒a=((((M/m)+1)g sin θ)/((M/m)+sin^2 θ))=−(ds^2 /dt^2 ) x_S =(1/(M+m))[Mx_M +m(x_M +s cos θ)] =(1/(M+m))[(M+m)x_M +ms cos θ] y_S =(m/(M+m))s sin θ ⇒a_(S,x) =(d^2 x_M /dt^2 )=(1/(M+m))[(M+m)A−ma cos θ) =(1/(M+m)){(((M+m)g sin θ cos θ)/((M/m)+sin^2 θ))−((m((M/m)+1)g sin θ cos θ)/((M/m)+sin^2 θ))} =0 (as expected) ⇒a_(S,y) =−(d^2 y_S /dt^2 )=(m/(M+m))a sin θ=((g sin^2 θ)/((M/m)+sin^2 θ))](Q111055.png)
$${N}\:\mathrm{sin}\:\theta={MA} \\ $$$${N}+{mA}\:\mathrm{sin}\:\theta={mg}\:\mathrm{cos}\:\theta \\ $$$${mg}\:\mathrm{sin}\:\theta+{mA}\:\mathrm{cos}\:\theta={ma} \\ $$$$\Rightarrow{g}\:\mathrm{sin}\:\theta+{A}\:\mathrm{cos}\:\theta={a} \\ $$$${A}\left(\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta\right)={g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{A}=\frac{{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}=\frac{{dx}_{{M}} ^{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}=\frac{\left(\frac{{M}}{{m}}+\mathrm{1}\right){g}\:\mathrm{sin}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}=−\frac{{ds}^{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$$ \\ $$$${x}_{{S}} =\frac{\mathrm{1}}{{M}+{m}}\left[{Mx}_{{M}} +{m}\left({x}_{{M}} +{s}\:\mathrm{cos}\:\theta\right)\right] \\ $$$$=\frac{\mathrm{1}}{{M}+{m}}\left[\left({M}+{m}\right){x}_{{M}} +{ms}\:\mathrm{cos}\:\theta\right] \\ $$$${y}_{{S}} =\frac{{m}}{{M}+{m}}{s}\:\mathrm{sin}\:\theta \\ $$$$ \\ $$$$\Rightarrow{a}_{{S},{x}} =\frac{{d}^{\mathrm{2}} {x}_{{M}} }{{dt}^{\mathrm{2}} }=\frac{\mathrm{1}}{{M}+{m}}\left[\left({M}+{m}\right){A}−{ma}\:\mathrm{cos}\:\theta\right) \\ $$$$=\frac{\mathrm{1}}{{M}+{m}}\left\{\frac{\left({M}+{m}\right){g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{{m}\left(\frac{{M}}{{m}}+\mathrm{1}\right){g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}\right\} \\ $$$$=\mathrm{0}\:\left({as}\:{expected}\right) \\ $$$$\Rightarrow{a}_{{S},{y}} =−\frac{{d}^{\mathrm{2}} {y}_{{S}} }{{dt}^{\mathrm{2}} }=\frac{{m}}{{M}+{m}}{a}\:\mathrm{sin}\:\theta=\frac{{g}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$
Commented by ajfour last updated on 01/Sep/20
$${Beautiful}\:{Sir},\:{understood};\:\mathcal{TH}{an}\mathcal{KS}! \\ $$