A solid sphere is rolling on a frictionless plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.

A solid sphere is rolling on a frictionless plane surface about its ax

1.	A solid sphere is rolling on a frictionless plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.
Answer» SOLUTION :Rot. `K.E.=(1)/(2) I OMEGA^(2)=(1)/(2)xx(2)/(4) mR^(2)xx(v^(2))/(R^(2)) "" ( "as" omega=(v)/(R), I=(2)/(5) mR^(2))` `=(1)/(5) mv^(2)` Total energy - Translational `K.E.+"Rot" K.E.=(1)/(2) mv^(2)+(1)/(5) mv^(2)=(7)/(10) mv^(2)` `("Rot. K.E.")/("Total Energy")=((1)/(5)mv^(2))/((7)/(10)mv^(2))=(2)/(7)`