[tex]\begin{gathered} \text{From the fre}e\text{ body diagram} \\ \uparrow\sum ^{}_{}Fy=-Ma \\ T1-Mg=-Ma\text{ (1)} \\ \text{Now, for the sphere} \\ T2\cdot R=I\alpha \\ I=\frac{2}{5}MR^2 \\ T2\cdot R=\frac{2}{5}MR^2\alpha \\ \text{also } \\ a=\alpha R \\ Solvi\text{ng }\alpha \\ \alpha=\frac{a}{R} \\ \text{This is possible because the }stri\text{ng }does\text{ not slip} \\ \text{Hence} \\ T2\cdot R=\frac{2}{5}MR^2\frac{a}{R} \\ T2\cdot R=\frac{2}{5}MRa \\ \text{Solving T2} \\ T2=\frac{2}{5R}MRa \\ T2=\frac{2}{5}Ma \\ \text{For the disk} \\ T2R-T1R=I\alpha \\ I=\frac{1}{2}MR^2 \\ \frac{2}{5}MaR-T1R=\frac{1}{2}MR^2\frac{a}{R} \\ \text{Solving T1} \\ \frac{2}{5}MaR-T1R=\frac{1}{2}MR^{}a \\ \frac{2}{5}Ma-T1=\frac{1}{2}M^{}a \\ T1=\frac{2}{5}Ma-\frac{1}{2}M^{}a \\ T1=\frac{-1}{10}Ma \\ \text{With T1 in (1)} \\ \frac{-1}{10}Ma-Mg=-Ma\text{ } \\ \text{SOlving a} \\ Mg=\frac{9}{10}Ma \\ a=\frac{10}{9}g \\ \text{The acceleration is} \\ a=\frac{10}{9}g \end{gathered}[/tex]
