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Thin spherical shell: I=(2/3)MR^2

Solid sphere: I=(2/5)MR^2

Thanks in advance.

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- Thread starter copperboy
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- #1

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Thin spherical shell: I=(2/3)MR^2

Solid sphere: I=(2/5)MR^2

Thanks in advance.

- #2

Tide

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[tex]I = \int r^2 dm[/tex]

In the case of the shell the element of mass is [itex]dm = M {dA} /{4 \pi R^2}[/itex] where [itex]dA = R^2 \sin \theta d\theta d\phi[/tex]. The distance to a point on the shell from the z-axis is [itex]R^2 \sin^2 \theta[/itex] so

[tex]I = \frac {M}{4 \pi R^2} \int_{0}^{2 \pi} d\phi \int_{-\pi /2}^{\pi /2}R^4 \sin^3 \theta d\theta[/tex]

from which the desired result follows.

In the case of the solid sphere you will work with a volume integral.

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Could you please explaim why [itex]dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?

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arildno

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Consider an area segment dA on a sphere with radius R.copperboy said:Could you please explaim why [itex]dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?

We approximate this with a rectangle:

a)Two of the sides are arclengths along great circles; the length of each of these is [tex]Rd\theta[/tex]

b) The other two are arclengths in THE HORIZONTAL PLANE; the local radius there is [tex]R\sin\theta[/tex]

Hence, the arclenth is [tex]R\sin\theta{d\phi}[/tex]

c) Multiplying together, we get:

[tex]dA=R^{2}\sin\theta{d\theta}d\phi[/tex]

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Tide

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copperboy said:Could you please explaim why [itex]dA = R^2 \sin \theta d\theta d\phi[/tex] in detail?

dA is a differential element of area on a spherical surface using spherical coordinates and it represents, to lowest order in differentials, the area of a rectangle [itex]R \sin \theta d\phi[/itex] high and [itex]R d\theta[/itex] units wide.

- #6

ruffneck2

Can someone please explain how and when to use these formulas.

Thank You.

- #7

siddharth

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You must be familiar with the equation F=ma.

We use the Moment of Inertia to find the Torque.

Like how the force is directly propotional to acceleration, Torque is directly propotional to angular acceleration, the propotionality constant being the moment of inertia.

Also remember about what axis you are taking the moment of Inertia while solving problems. In most problems i have encounterd which ask you to find the angular acceleration, first find the torqure taking the vector product R X F. Next find the moment of inertia of the object about the required axis. Then the acceleration can be found

Also, for a solid sphere, perphaps deriving the moment of inertia by intergrating thin rings would be easier than a volume intergral?

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