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17.A solid cy linder rolls down an inclined plane. Its mass is 2 kg and radius 0.1 m. if the height of the inclined plane is 4m, what is its rotational K.E. when it reaches the foot of the plane?126.13 J

Answer»

Mgh = KE of translation + KE of rotationMgh= 1/2 Mv² + 1/2 Iω²Mgh = 1/2 M(ωR)² + 1/2 (1/2 MR²)ω²Mgh = 3/4 MR²ω²

KE of rotation = K= 1/2 Iω²= 1/2 (1/2 MR²) 4gh /3R²K= Mgh/ 3= 2 x 9.8 x 4/ 3=26.13 J



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