A ring of mass `m` is rolling without slipping with linear speed `v` as shown in figure. Four particles each of mass `m` are also attached at points A, B , C and D find total kinetic energy of the system.

Explanation

Earlier we have have learned that in case of pure rolling,
`omega=(v)/(R),v_(A)=0,v_(B)=v_(D)=sqrt(2)v` and `v_(C)=2v`
Now, total kinetic energy `=`[ tanslational kinetic energy of four particles`]+[` translational kinetic energy of ring `+` rotational kinetic energy of ring]
`therefore(KE)_("total")=[(1)/(2)m(0)^(2)+(1)/(2)m(sqrt(2)v)^(2)+(1)/(2)m(sqrt(2)v)^(2)+(1)/(2)m(2v)^(2)]+[(1)/(2)mv^(2)+(1)/(2)Iomega^(2)]`
Substituting `I=mR^(2)` and `omega=(v)/(R)`
we get `(KE)_("total")=5mv^(2)`