Question 1

I was doing fairly fine until I came across this question, I know that linear acceleration equals R(alpha) and I tried arranging the force diagram but I am a bit lost...

I was doing fairly fine until I came across this question, I know that linear acceleration-example-1

Question 2

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

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