# Why is the moment of inertia for a hollow sphere higher than a uniform sphere?

I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.

-
A simpler question: why does a ring have a higher moment of inertia than a disk? –  Beta Feb 23 '14 at 20:00
To expound Beta's comment: What has higher moment of inertia: A uniform bar, or a massless bar with two point weights at it's ends? –  nbubis Feb 23 '14 at 20:47
Because more of its mass is located outwards. –  ja72 Feb 24 '14 at 0:19
do the spheres have the same mass or are they made out of the same material? The answer will be different for each option –  ratchet freak Feb 24 '14 at 9:53
The simple intuitive understanding: imagine (A) swinging around you a 1kg metal ball, on a string a meter long. Versus (B) swinging a 1kg, 1 meter, metal bar. Of course (A) has much more zoomf. (To use a technical word. :) ) The sphere example is just the same. –  Joe Blow Feb 26 '14 at 12:26

The key is... the closest the mass to the axis of rotation, the easiest to add angular velocity to the body.

For instance a figure skater rotates faster when she puts her limbs closer to her body.

Let's see how it works from a more intuitive fashion:

For instance, in the figure bellow, trying to lift up the table (A) would be easier compared with the table (B).

In both cases the mass of each individual box is the same, but in (A) you have a better leaver by means of the distance from the border where the force is being applied, to each box.

Therefore, the table (B) would be harder to lift up, even when R (length of the table) and M (total mass of the four boxes) are the same.

Now let's see how it works in the case with the spheres:

1. Lets make the sphere a disk, and then divide it in slides.
2. Make fixed the center of mass of the disc and move all the slides to one side.
3. Now we have a similar scenerario to the one with the tables.

Both spheres, the solid and hollow one, rotate around their center of mass in the same way that the table rotates around the legs at the opposite side where the force is being applied.

To make sence of the step 2 where the mass of all slides is collapsed into one slide, think on a Merry-Go-Round where all the kids move to one side keeping fixed their distance to the axis of rotation.

-
I think the cause of the confusion is that, intuitively, his example is symmetric while this one clearly isn't. –  Adrian Petrescu Feb 26 '14 at 19:33
@AdrianPetrescu Thanks for pointing it out, I've edited the answer to make the connection clear, I hope. –  rraallvv Feb 26 '14 at 21:18

A hollow sphere will have a much larger moment of inertia than a uniform sphere of the same size and the same mass.

If this seems counterintuitive, you probably carry a mental image of creating the hollow sphere by removing internal mass from the uniform sphere. This is an incorrect image, as such a process would create a hollow sphere of much lighter mass than the uniform sphere. The correct mental model corresponds to moving internal mass to the surface of the sphere.

-
On the other hand, if you start with a solid uniform sphere, and you do remove a smaller (concentric) sphere from its interior, then you end up with a "thick" spherical shell whose mass and moment of inertia are both smaller than those of the original solid ball. The intuition that says "if I remove something, the moment gets smaller" is right. –  Jeppe Stig Nielsen Feb 24 '14 at 9:23
The mental image is incorrect in the sense that it doesn't correspond to creating a shell of the same mass as the uniform sphere. –  Johannes Feb 24 '14 at 17:27

Here's an illustration of a uniform sphere and a hollow sphere mid-sections with the same mass, if you better understand these things visually:

-
Nice illustration, but how does it answer the question? –  Paŭlo Ebermann Feb 25 '14 at 18:59
@PaŭloEbermann visually? –  user1306322 Feb 25 '14 at 19:34
This isn't rigorous, you haven't defined "low" and "high" and the scale of the gradient! –  Thomas Feb 26 '14 at 11:34
@Thomas feel free to improve upon my illustration or to post your own vision of it, because, frankly, I have little idea of what you're talking about. I just figured out how it should look and drew it. –  user1306322 Feb 26 '14 at 12:47
@user130632 Never mind. –  Thomas Feb 26 '14 at 12:48

The moment of inertia of a body about an axis is a measure of how far the mass is distributed from that point. For a solid sphere of mass $m$, radius $r$, you have the mass distributed continuously from the center to the radius. However, for a hollow sphere of mass $m$, inner radius $r_i$ and outer radius the same as before, $r$, you have all the mass slightly farther from the center.

Since the entire mass is farther from the center, it is harder to change its angular momentum, and its moment of inertia is larger

-

The question is not why - that is misleading. It is when as under certain conditions only is this true. The moment of inertia is proportional to the mass and distribution of that mass about the axis.

-

Because perpendicular force x radius is a measure of angular momentum, and all the AM of different point masses added up to give the total effective AM.

So the larger the radius of the masses the better, in this case we can arrange all of them to reside as point masses at the edge of the circle (because rotation is the motion of action).

-

If you consider the sphere as two hemispheres, then the centre of gravity of each is 3/8r from the centre. (http://my.safaribooksonline.com/book/mechanical-engineering/9789332503489/6-centroid-and-moment-of-inertia/chap6_sub14_xhtml) The hollow sphere, when considered as two thin shelled hollow hemispheres has a c. of g. at r/2 from the centre. http://my.safaribooksonline.com/book/mechanical-engineering/9789332503489/6-centroid-and-moment-of-inertia/chap6_sub15_xhtml Hence if they are the same size and mass (and therefore a more dense material for the hollow body), the hollow hemispheres have their effective mass further from the fulcrum.

-

The moment of inertia of a hollow sphere would be higher than a solid sphere of equal radius, only if the unmentioned assumption (same mass) is true! This is typically untrue, because of another assumption, that the hollow and solid spheres (of equal radius) are made of the same density material.

If they are made of the same density material, there is no way that the hollow sphere will have the same mass as the solid one (less volume). If the spheres are made from the same density material, the solid one will have the higher moment of inertia, because of its larger mass.

-
… because the difference of a solid ball and a sphere of the same exterior radius consists of a solid ball of the same dimensions as the sphere’s cavity. The “larger mass” argument proves nothing. –  Incnis Mrsi Oct 24 '14 at 17:41

## protected by Qmechanic♦Feb 26 '14 at 22:54

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.