A circular disc rolls down an inclined plane without slipping. What fraction of its total energy is translational ?

The kinetic energy of translational motion is given by( K.E. )t=12mv2=12mr2ω2The ratational kinetic energy is given by(K.E.)r=12Iω2=1212mr2ω2=14mr2ω2∴  Total (K.E.) r=12mr2ω2+14mr2ω2=34mr2ω2The ratio of translational kinetic energy to total kinetic energy is given by( K.E. )t( K.E. )T=(1/2)mr2ω2(3/4)mr2ω2=23