Two beads each of mass m are welded at the ends of two light rigid rods each of length l. If the pivots are smooth then find the ratio of translational and rotational kinetic energy of system.

We know translational kinetic energy Kt=12mvC2 and rotational kinetic energy Kr=12ICω2Distance of centre of mass from pivot end,rC=ml+m(2l)m+m=32lMoment of inertia of system about centre of mass,IC=ml22+ml22=ml22Angular velocity of rod ω=v2lvelocity of centre of mass, vcm=ω32l=v2l32l=34vTranslational kinetic energy of system,Ktransiational =12Mtotal ⋅vcm2=12(2m)34v2=916mv2Kroltaional =12ICMω2=12ml22v2l2=116mv2This gives KtKr=9.