Question

kinetic energy in rolling motion its derivation and statement​

Answer 1

Kinetic Energy for Rolling Motion

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

K.E = KT + KR

= 1/2 mv² + 1/2 Iω²

= 1/2 m Vcm² + 1/2 Iω²

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

= 1/2 m Vcm² + 1/2 mk² ω²

For rolling without slipping the mathematical condition is rω = Vcm

K = 1/2 m² ω² + 1/2 mk² ω²

= 1/2 m Vcm² + 1/2 mk² \( \frac{v_cm²}{r²} \)

⇒ K = 1/2 m Vcm² [ 1 + k²r² ]

This is the kinetic energy of a rolling motion.