A string is wound around a disc of radius r and mass M and at the free end of the string a body of mass m is suspended. The body is then allowed to descend. Show that the angular accelerationof the disc = `(mg)/(R(M+(m)/(2)))`

Explanation

The suspended body and the forces acting on the disc are shown in fig.
The equation of the linear motion of the suspended body is `ma=mg-T`
(where T = tension in the string)
`therefore T=m(g-a)`
Now torque acting on disc `tau=RT [because vectau=vecrxxvecF]`
`therefore Ialpha=RT`
`therefore alpha=(RT)/(I)=(Rm(g-a))/(I)`
`therefore alpha=(Rm(g-a))/(M(R^(2))/(2))[because I=(MR^(2))/(2)]`
`therefore alpha=(2m)/(RM)(g-a)" but "a=Ralpha`
`therefore alpha=(2mg)/(RM)-(2mRalpha)/(RM)`
`therefore alpha+(2mRalpha)/(RM)=(2mg)/(RM)`
`therefore alpha(1+(2m)/(M))=(2mg)/(RM)`
`therefore alpha=(2mg)/(RM(1+(2m)/(M)))=(mg)/((R)/(2)(M+(2mM)/(M)))=(mg)/(R((M)/(2)+m))`