Rigid body motion - thin disc

[Celso]

Why is the gravitational potential energy of the chain's center of mass equal to the total kinetic energy of the disc after it was fully wrapped? My first thought was to write $E_{0}=(M/2+M)g∗2πR=E_{f}= Ep$ (from the chain) $+Ec$ (from the disc). Instead he wrote $mg \frac{l}{2}$ = $\frac{1}{2} I \omega² + \frac{1}{2} M v²$

 

[Andrew Mason]

Summary: This is an example from the book (Analytical mechanics, Fowles), I don't get the energy equation step

$E_{rot} = 1/2 I \omega²$
$E_{translational} = 1/2 mv²$
$E_{potential} = mgh$

Why is the gravitational potential energy of the chain's center of mass equal to the total kinetic energy of the disc after it was fully wrapped? My first thought was to write $E_{0}=(M/2+M)g∗2πR=E_{f}= Ep$ (from the chain) $+Ec$ (from the disc). Instead he wrote $mg \frac{l}{2}$ = $\frac{1}{2} I \omega² + \frac{1}{2} M v²$

The decrease in the chain's gravitational potential energy is equal to the (rotational) kinetic energy of the disc PLUS the (translational) kinetic energy of the chain. So, as the author wrote,:

mg \frac{l}{2} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2 [Note: m not M in the last term]

AM