A disc is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies ?

Options

A. `1 :3`
B. `3:1`
C. `1:2`
D. `2:1`

Explanation

The rotational kinetic energy of the disc is
`K _(omega ) = 1/2 I omega a^(2) = 1/2 ((1)/(2) MR ^(2)) omega ^(2)`
`= (1)/(4) MR ^(2) omega ^(2)`
The transitional kinetic energy is
`K _("mass" ) = 1/2 Mv _(CM ) ^(2)`
Where, `v _(CM)` is the linear velocity of its centre of mass Now, `v _(CM) = R omega` (for pure rolling)
Therefore, `K _("mass") = (1)/(2) M R ^(2) omega ^(2)`
Thus, `K _("total") = (1)/(4) MR ^(2) omega ^(2) + (1)/(2) MR ^(2) omega ^(2)`
`= (3)/(4) MR ^(2) omega^(2)`
`therefore (K _(omega ))/( K _("total ") ) = ((1)/(4) MR ^(2) omega ^(2))/( (3)/(4) MR ^(2) omega ^(2)) = (1)/(3)`