A disc is undergoing pure rolling motion on a flat surface. The radius of the disc is 5cm & velocity of centre of mass is 10m/s. What is the ratio of kinetic energy as seen from the ground frame to a frame moving along the disc with a speed of 10 m/s?

3:1

The kinetic energy of a pure rolling disc as seen from the ground frame is given by: 1/2MV2 + 1/2Iw2.

I = MR²/2 w = V/R

∴ 1/2MV² + 1/2Iw² = 1/2MV² + 1/4MV² = 3/4MV².

Kinetic energy as seen from the moving frame will only be the energy due to rotation about the disc’s axis.

∴ Kinetic energy from moving frame = 1/2Iw² = 1/4MV².

∴ Ratio = (3/4MV²) / (1/4MV²)

= 3:1.