A uniform disc of mass m is fitted (pivoted smoothly) with a rod of mass $\frac{\mathrm{m}}{2}$. If the bottom of the rod is pulled with a velocity v, it moves without changing its angle of orientation and the disc rolls without sliding. Thd kinetic energy of the system (rod + disc) is.

 

 

Rolling can be considered as pure rotation about point of contact

${\mathrm{K}}_{\mathrm{rolling}} = \frac{1}{2}{\mathrm{I}}_{\mathrm{P}}{\mathrm{\omega }}^2 = \frac{1}{2}(\frac{{\mathrm{mR}}^2}{2}+{\mathrm{mR}}^2){\mathrm{\omega }}^2 = \frac{3}{4}{\mathrm{mR}}^2{\mathrm{\omega }}^2$-----------(i)

The rod translates with the velocity v hence velocity of centre of disc will also be v

$\mathrm{v} = \mathrm{\omega R}------\left(\mathrm{ii}\right)$

From (i) and (ii); ${\mathrm{K}}_{\mathrm{rolling}} = \frac{3}{4}{\mathrm{mv}}^2$

Kinetic energy of the rod ${\mathrm{K}}_{\mathrm{rod}} = \frac{1}{2}\left(\frac{\mathrm{m}}{2}\right){\mathrm{v}}^2 = \frac{{\mathrm{mv}}^2}{4}$-------(iii)

${\mathrm{K}}_{\mathrm{total}} = {\mathrm{K}}_{\mathrm{rolling}}+{\mathrm{K}}_{\mathrm{rod}} = {\mathrm{mv}}^2$