物理,SMALL BLOCKS,MOMENT

用戶91254

2013-06-20 6:49 am

Small blocks, each with mass m , are clamped at the ends and at the center of a
rod of length L and negligible mass.

2. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end.

Express your answer in terms of the given quantities.

我知道答案係11/16mL^2,但我唔明白點解唔係m(3/4*L)^2+m(1/4*L)^2=10/16mL^2

請高手解答!

回答 (1)

天同

2013-06-20 7:37 am

✔ 最佳答案

I suppose there are 3 small blocks, two are placed each at the two ends of the rod, and the third one is placed at the centre.

Thus, one block (at the near end) is at distance of L/4 from the axis, the block at the far end is (3L/4) from the axis, and the central block is L/4 from the axis.

Moment of inertia of the block at the near end
= m(L/4)^2 = mL^2/16
Moment of inertia of the block at the far end
= m(3L/4)^2 = 9mL^2/16
Moment of inertia of the block at the centre
= m(L/4)^2 = mL^2/16

Hence, total moment of inertia
= mL^2/16 + 9mL^2/16 + mL^2/16
= 11mL^2/16

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