Question

A ring and a disc rolling on a horizontal surface without slipping with same linear velocity

Answer 1

your question is incomplete. A complete question is ---> a ring and a disc of sane mass roll without slipping along a horizontal surface with same velocity. if the kinetic energy of ring is 8j, then that of disc is ?

solution : moment of inertia of ring about its axis , I = mr²

moment of inertia of disc about its axis , I' = 1/2 mr²

now, kinetic energy of ring , K.E = 1/2 Iw² + 1/2 mv²

where, w is angular frequency.

so, w = v/r

now, K.E = 1/2 (mr²) × (v/r)² + 1/2mv²

or, 8 J = 1/2 × mr² × v²/r² + 1/2mv²

or, 8 = mv² ...........(1)

kinetic energy of disc , K.E' = 1/2 I' w² + 1/2 mv²

= 1/2 (1/2 mr²) × (v/r)² + 1/2 mv²

= 1/4 × mr² × v²/r² + 1/2mv²

= 1/4 mv² + 1/2 mv²

= 3/4 mv²

from equation (1),

= 3/4 × 8 = 6J

hence, kinetic energy of disc is 6J

Answer 2

Answer:

your answer is above mentioned attachment