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+(2I)/r^(2))`
B. `((M-m)g)/(M+m-(2I)/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_(1)) = Ialpha`………..(ii)
`Mg-T_(3)=Ma`………(iii)
`r(T_(3)-T_(2)) = Ialpha`………(iv)
and a =Ra

From eqs. (ii) and (iv), we get, `T_(3)-T_(1) = (2Ia)/r^(2)`
From Equs. (i) and (iii), (M-m)g = (M+m)a `+ T_(3)-T_(1)`
`(M-m)g = (M+m)a + (2Ia)/r^(2) rArr a = ((M-m)g)/(M+m+(2r)/r^(2)) rArr a=((M-m)g)/(M+m + (2I)/r^(2))`