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?
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126
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
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
Answered by
6
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:
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