A uniform cylinder of radius `r` and mass `m` can rotate freely about a fixed horizontal axis. A thin cord of length l and mass `m_(0)` is would on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length `x` of the hanging part of the end. the wound part of the cord is supposed to have its centre of gravity on the cylinder axis is shown in figure.

Explanation

Figure shows the free body diagram of the two bodies. Viz., cylinder and overhnging portion of the cord.
If `alpha` is the angular acceleration of the cylinder then from `tau=Ialpha`.
we have `Tr=[(mr^(2))/2+m_(0)-m_(0)/l(l-x)r^(2)]alpha`

`T=[m/2+m_(0)-(m_0x)/l]ralpha`....i
If `a` be the acceleration of the hanging part, the from `F=ma`.
`(m_(0)xg)/l-T=(m_(0)x)/l a=(m_(0)x)/l(alphar) `..........ii
[Linear acceleration of the rim of the cylinder is same as that of the cord]
Adding eqn i and ii we get
`(m_(0)x)/lg=alpha[m/2+m_(0)]rimpliesalpha=(m_(0)xg)/(rl[m/2+m_(0)])`
Alternate method: Taking the cylinder and the cord as a single system moving in a gravitational, field, its total energy (mechanical) should remain constant
Assuming the level of the cylinder's centre as datum or refeence level , we have `(dE)/(dt)=0`

i.e., `d/(dt)[1/2((mr^(2))/2)(v/r)^(2)+1/2m_(0)v^(2)-(m_(0)x)/1g(x/2)]=0`
[Here `v` is the speed of the hanging part]
`implies(m/2+m_(0))2v((dv)/(dt))-(m_(0)g)/1(2x(dx)/(dt))=0`
`implies (m_(0)+m/2)(dv)/(dt)=(m_(0)gx)/1implies(dv)/(dt)=(m_(0)gx)/(l(m/2+m_(0)))`
Therefore, the angular acceleration of the cylinder
`=alpha/r=(m_(0)gx)/(rl(m/m_(0)))`