A uniform disc of mass `m` is fitted (pivoted smoothly) with a rod of mass `m//2`. If the bottom of the rod os pulled with a velocity `v`, it moves without changing its orientation and the disc rolls without sliding. The kinetic energy of the system (rod + disc) is.
.

Explanation

Rolling can be considered as pure rotation about point of contact
`K_("rolling")=1/2I_(P)Iomega^(2)`
`=1/2((mR^(2))/2+mR^(2))omega^(2)=3/4mR^(2)omega^(2)`……..i
The rod translates with the velocity `v` hence velocity of centre of disc will also be `v`
`V=omegaR` ............ii
Form i and ii `K_("rolling")=3/4mv^(2)`
Kinetic energy of the rod `K_("rod")=1/2(m/2)v^(2)=(mv^(2))/4`.........iii
`K_("total")=k_("rolling")+k_("rod")=mv^(2)`