Question

A solid cylinder rolls down an inclined plane. Its mass is 2kg and radius is 0.1m. If the height of the inclined plane is 4m, what is its rotational K.E. when it reaches 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

Answer 2

Answer: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

Explanation: