Shown in the diagram is a system of two bodies, a block of mas `m` and a disc of mass `M`, held in equilibrium. If the string `3` is burnt, find the acceleration of the disc. Neglect the masses of the pulleys `P` and `Q`. (The coefficient of friction between the block and horizontal surface is `mu`)

Explanation

Referring to the free-body diagram we write the equation of motion.
For the block `m` :
`T_(1)-muN=ma_(1)`……(`1`)
and `N-mg=0`…(`2`)
For pulley `P` :
`m_(p)g+T_(2)-2T_(1)=m_(p)a_(p)`
Since the pulley is light,
`:. m_(p)~~0`
`rArr T_(2)=2T_(1)`.....(`3`)
For disc `M`.
`Mg-2T_(2)=Ma_(2)`......(`4`)
From the geometry of the figure the positions of the parts `m`, `P` and `M` are related as `x_(1)+2y_(1)=l_(1)` and `(y_(2)-y_(1))+y_(2)=l_(2)`, where `l_(1)` and `l_(2)` are the lengths of the strings `1` and `2`, respectively. Eliminating `y_(1)` from these two equations, we obtain,
`x_(1)+4y_(2)=l_(1)+2l_(2)` ,
Differntiating both sides twice w.r.t. time, we obtain `(d^(2)x_(1))/(dt^(2))+4(d^(1)y_(2))/(dt^(2))=0`
`rArra_(1)=4a_(2)`.....(`5`)
From Eqs. (`1`) and (`2`), we get
`T_(1)-mumg=ma_(1)`.......(`6`)
From Eqs. (`3`) and (`4`), we get
`Mg-4T_(1)=Ma_(2)`.....(`7`)
Eliminating `T_(1)` from Eqs. (`6`) and (`7`) and putting `a_(1)=4a_(2)`, we obtain
`a_(2)=(M-4mum)/(M+16m)g`.