The pulleys in figure are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Options

A. `((M-m)g)/((M+m+(2l)/r^(2)))`
B. `((M-m)g)/((M+m-(2l)/r^(2)))`
C. `((M-m)g)/((M+m+I/r^(2)))`
D. `((M-m)g)/((M+m-I/r^(2))`

Answer

A

Explanation

`T_(1)-mg=ma` ………….i
`r(T_(2)-T_(2))=Ialpha`…….ii

`Mg-T_(3)=Ma`………iii
`r(T_(30-T_(2))=Ialpha`……….iv
and `a=Ralpha`
From eqn ii and iv we get
`(M-m)g=(M+m)a=T_(3)-T_(1)`
`(M-m)g=(M-m)a+(2Ia)/(r^(2))`
`implies a((M-m)g)/((M+m+(2l)/(r^(2))))`