MoI of a Sphere using Spherical Coordinates

In summary, the conversation discusses the calculation of the moment of inertia of a uniformly distributed sphere about an axis through its center. The equations used in the attempt at a solution involve the mass, radius, and density of the sphere. However, the result obtained does not match the expected result of ##\frac{2}{5}MR^{2}##. The error is identified as the use of spherical coordinates and the lack of accounting for the perpendicular distance from the axis of rotation to a given mass element. The correct solution is obtained by using the perpendicular distance formula in the moment of inertia integral.
  • #1
Ackbach
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Homework Statement



Calculate the moment of inertia of a uniformly distributed sphere about an axis through its center.

Homework Equations



I know that
$$I= \frac{2}{5} M R^{2},$$
where ##M## is the mass and ##R## is the radius of the sphere. However, for some reason,
when I do this integration in spherical coordinates, I do not get this result.

The Attempt at a Solution



The density is
$$ \rho= \frac{M}{V}= \frac{3M}{4 \pi R^{3}}.$$
Then the moment of inertia is
$$I= \int r^{2} \,dm= \rho \iiint_{V}r^{2} \,dV
= \rho \int_{0}^{2 \pi} \int_{0}^{ \pi} \int_{0}^{R} r^{4} \sin(\theta) \, dr \, d\theta \, d\phi
=4 \pi \rho \frac{R^{5}}{5}
$$
$$=4 \pi \frac{3M}{4 \pi R^{3}} \frac{R^{5}}{5}= \frac{3}{5} MR^{2}.$$
Where is my error?

I know, I know: all the derivations of this result I can find split the sphere up into
disks. I understand those derivations: I want to know where this one is wrong.
 
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  • #2
The ##r## in the moment of inertia integral is the perpendicular distance from the axis of rotation to a given mass element ##dm##. The ##r## in spherical coordinates is the distance to ##dm## from the center of the 2-sphere. The two only agree for mass elements that lie on the equatorial plane.
 
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  • #3
If you really want to use spherical coordinates, then note that the perpendicular distance from the axis of rotation to a given mass element will be given by ##r_{\perp } = r\sin\theta## hence the moment of inertial integral becomes ##I = \rho \int _{S^{2}}r^{2}\sin^{2}\theta dV = \frac{3}{10}MR^{2} \int_{0}^{\pi}\sin^{3}\theta d\theta = \frac{3}{10}\frac{4}{3}MR^{2} = \frac{2}{5}MR^{2}##.
 
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1. What is the formula for finding the moment of inertia of a sphere using spherical coordinates?

The formula for finding the moment of inertia of a sphere using spherical coordinates is I = 2/3 * m * r2, where m is the mass of the sphere and r is the radius.

2. How does the moment of inertia change with respect to the radius of the sphere?

The moment of inertia increases with the square of the radius. This means that as the radius of the sphere increases, the moment of inertia also increases.

3. Can the moment of inertia of a sphere change if its mass is distributed differently?

Yes, the moment of inertia can change depending on how the mass is distributed within the sphere. For example, if the mass is more concentrated towards the center of the sphere, the moment of inertia will be smaller compared to if the mass is evenly distributed throughout the sphere.

4. How does the moment of inertia of a sphere differ from that of other shapes?

The moment of inertia of a sphere is unique because it is the same for any rotation axis passing through the center of the sphere. This is not the case for other shapes, where the moment of inertia can vary depending on the axis of rotation.

5. Can the moment of inertia of a sphere be negative?

No, the moment of inertia of a sphere cannot be negative. It is always a positive value, as it represents the resistance of an object to changes in its rotational motion.

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