A wheel of mass `5 kg` and radius `0.40 m` is rolling on a road without sliding with angular velocity `10 rad s^-1`. The moment of ineria of the wheel about the axis of rotation is `0.65 kg m^2`. The percentage of kinetic energy of rotate in the total kinetic energy of the wheel is.

Explanation

Here, `m = 5 kg, r = 0.4m , omega = 10 rad//s`,
`I =0.65 kg m^(2) , v = r omega = 0.4 xx 10 = 4.0 m//s`
Translational `KE = (1)/(2)mv^(2) = (1)/(2) xx 5 xx (4.0)^(2) = 40 J`
Rotational `KE = (1)/(2)I omega^(2) = (1)/(2) xx 0.65 (10)^(2) = 32.5 J`
Total `KE =` Translational `KE +` Rotational `KE`
`= 40 + 32.5 = 72.5 J`
`(KE "of translational")/("Total" KE) = (40)/(72.5) = 0.552 =55.2%`