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)))

A string is wound around a disc of radius r and mass M and at the free

1.	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)))
Answer» Solution : 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))`