For the system shown in figure, `M=1kg` `m=0.2` kg, `r=0.2m` calculate `(g=10m//s^(2))`
(a). The linear acceleration of hoop,
(b). The angular acceleration of the hoop of mass `M` and
(c). The tension in the rope.

Explanation

If `alpha` be the angular acceleration of the hoop and `a` be the acceleration if its centre, accelaration of `m` would be `alphar+a`
Here `Tr=Ialpha` [where `I=` moment of inertia of the hoop about the horizontal axis passing through its centre].
Also, `T=Ma` and `mg-T=m[a+ar]`
solving we get
`a=(mg)/([M=2m])=(2)/(1.4)=1.43m//s^(2)`
Hence, `T=1.43N`
and `alpha=(Tr)/(I)=(T)/(Mr)=7.15rad//s^(2)`