A uniform solid cylinder of mass 'm' can rotate freely about its axis which is kept horizontal. A particle of mass m0 hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is
Suppose that the tension in the string is T.
Then, =m0g−T=m0a
∴T=m0(g−a)
where a = acceleration
Further, T.r = I α
α = angular acceleration and T.r = moment of force acting on cylinder
or T=Iαr
=(mr22)(αr)=mrα2=ma2
∴a=2Tm=2m0(g−a)m
Solving for a, we get
a=(2m0gm+2m0).