can you answer the question in this link? thx
http://i649.photobucket.com/albums/uu213/cyn001/16.jpg
a rotational problem
2009-12-05 2:03 am
回答 (2)
2009-12-05 3:24 am
✔ 最佳答案
a. By parallel axis theorem,moment of inertia of the combination about O, I
= [1/3 Ml^2 + M(l/2)^2] + [1/3 Ml^2 + M(3l/2)^2] + ml^2 + m(2l)^2
= (19M/6 + 5m)l^2
b. Kinetic energy of rotation about O
= 1/2 Iw^2
= 1/2 (19M/6 + 5m)l^2w^2
參考: Physics king
2009-12-05 7:37 am
(a) moment of inertia of rod with length 2l about O = M(2l)^2/3 = (8/3)Ml^2
moment of inertia of the two masses about O
= ml^2 + m(2l)^2 = 5ml^2
Hence, total moment of inertia
= [(8/3)Ml^2 + 5ml^2] = [8M/3 + 5m].l^2
(b) kinetic energy = (1/2)[8M/3 + 5m].l^2.w^2
moment of inertia of the two masses about O
= ml^2 + m(2l)^2 = 5ml^2
Hence, total moment of inertia
= [(8/3)Ml^2 + 5ml^2] = [8M/3 + 5m].l^2
(b) kinetic energy = (1/2)[8M/3 + 5m].l^2.w^2
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091204000051KK00917