Question

A solid cylinder of mass 2 kg and radius 0.1 meter roll down and incline plane of height 3 meter calculate its rotational energy when it reaches the foot of the plane​

Answer 1

Answer:

=26.13 J

Step-by-step explanation:

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