Question

Solid cylinder of mass 2 kg and radius 0.1 m rolls down an inclined plane of height 3m. Calculate its rotational energy when it reaches the foot of the plane.

Answer 1

Mgh = KE of translation + KE of rotation 

Mgh= 1/2 Mv² + 1/2 Iω²

Mgh = 1/2 M(ωR)² + 1/2 (1/2 MR²) ω²

Mgh = 3/4 MR²ω²

KE of rotation = K= 1/2 Iω²

= 1/2 (1/2 MR²) 4gh /3R²

K= Mgh/ 3

= 2 x 9.8 x 4/ 3

=26.13 J