Question

Obtained an expression for the total kinetic energy of a rolling of a rolling body in the from 1/2 mv² (1+k²/R²).

Answer 1

Kinetic Energy for Rolling Motion

Since rolling is the combination of rotational motion and translational motion,

K.E = KT + KR

= 12 mv² + 12 Iω²

= 12 m vcm² + 12 Iω²

I = moment of inertia. We can write moment of inertia as I = mk², k = radius of gyration.

= 12 m vcm² + 12 mk² ω²

For rolling without slipping the mathematical condition is rω = vcm

K = 12 m² ω² + 12 mk² ω²

= 12 m vcm² + 12 mk² \( \frac{v_cm²}{r²} \)

⇒ K = 12 m vcm² [ 1 + k²r² ]

This is the kinetic energy of a rolling motion.

Answer 2

rolling energy/total energy = translation kinetic energy + rotational kinetic energy
= 1/2mv2+ 1/2mv2+k2/R2


..since rotational kinetic energy=1/2Iw2