Roman C. answered • 11/07/16

Masters of Education Graduate with Mathematics Expertise

^{2}/2 independent of the angle. This can be seen as follows.

Samagra G.

asked • 11/07/16We have to find moment of inertia of a semi circular disc whose axis of rotation is passing through the centre making an angle k with the diameter and lying in the plane of disc .

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Roman C. answered • 11/07/16

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The answer is I=mr^{2}/2 independent of the angle. This can be seen as follows.

1. This is easily found to be the answer if k=0.

2. If k≠0, then the axis slices the semicircular disks into two sectors. If we take one sector and rotate it 180° in the plane and about the center, that piece's moment of inertia is preserved by symmetry. But then the new shape is the same disk with diameter on the axis of rotation. Thus the moment of inertia is the same as in case 1.

Steven W. answered • 11/09/16

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Physics Ph.D., *professional*, easygoing, 6000+ hours tutoring physics

The moment of inertia, in this situation, depends only on the total mass and how the mass is distributed around the axis, defined by:

I = ∫r^{2} dm

where r is the perpendicular distance of each differential unit of mass from the axis of rotation. If you take a "pie slice" of the kind Roman talked about, and rotate it 180^{o}, each differential unit of mass still has the same perpendicular distance from the axis. Hence, nothing about the calculation of I changes, because the mass is still distributed the same way, in terms of perpendicular distances, from the axis.^{}

You can also do the integral (I would recommend polar coordinates). You should find that the dependence of I on k drops out when the integral is fully evaluated.

Samagra G.

Can you please tell me a start how to use polar coordinates .

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11/09/16

Steven W.

tutor

The way I looked at it was to define the perpendicular distance from the axis as "rsinθ." And each little differential mass unit, dm, is just the (constant) mass density of the disc (which I will call μ) times the differential unit of area; which, in polar coordinates, is rdrdθ. Then the integral becomes:

∫(rsinθ)^{2}(μrdrdθ)

I integrated this in r from 0 to R, where R is the radius of the disc, and from -k to 180-k in θ. I lost a factor of two in there, but the k dependence did, indeed, drop out, as Roman's elegant solution states.

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11/09/16

Samagra G.

Why the answer using this is coming the half of original answer .

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11/09/16

Steven W.

tutor

I would have to check to see where the factor of 2 comes from. But the principle Roman expressed is absolutely correct, that there is no k dependence.

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11/09/16

Samagra G.

Yes that expression is correct

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11/09/16

Steven W.

tutor

Yes, actually, it should be MR^{2}/4, because the full disk would be MR^{2}/2.

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11/09/16

Samagra G.

Thanks a lot

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11/09/16

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Samagra G.

11/08/16