Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? 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.
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6$\begingroup$ A simpler question: why does a ring have a higher moment of inertia than a disk? $\endgroup$– BetaCommented Feb 23, 2014 at 20:00
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6$\begingroup$ 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? $\endgroup$– Nathaniel BubisCommented Feb 23, 2014 at 20:47
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11$\begingroup$ Because more of its mass is located outwards. $\endgroup$– John AlexiouCommented Feb 24, 2014 at 0:19
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$\begingroup$ do the spheres have the same mass or are they made out of the same material? The answer will be different for each option $\endgroup$– ratchet freakCommented Feb 24, 2014 at 9:53
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$\begingroup$ 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. $\endgroup$– FattieCommented Feb 26, 2014 at 12:26
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8 Answers
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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.
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5$\begingroup$ 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. $\endgroup$ Commented Feb 24, 2014 at 9:23
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3$\begingroup$ 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. $\endgroup$– JohannesCommented Feb 24, 2014 at 17:27
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The key is... the closer the mass to the axis of rotation, the easier it is 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 in a more intuitive fashion:
For instance, in the figure bellow, trying to lift up table (A) would be easier than table (B).
In both cases the mass of each individual box is the same, but in (A) you have a better lever because of the distance from the border where the force is being applied, to each box.
Therefore, 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:
- Let's make the sphere a disk, and then divide it in pieces.
- Make the center of mass of the disc fixed, and move all the pieces to one side.
- 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 to where the force is being applied.
To make sense of step 2, where the mass of the all pieces is collapsed, think on a Merry-Go-Round where all the kids move to one side keeping their distance to the axis of rotation fixed.
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$\begingroup$ I think the cause of the confusion is that, intuitively, his example is symmetric while this one clearly isn't. $\endgroup$ Commented Feb 26, 2014 at 19:33
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$\begingroup$ @AdrianPetrescu Thanks for pointing it out, I've edited the answer to make the connection clear, I hope. $\endgroup$– rraallvvCommented Feb 26, 2014 at 21:18
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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:
answered Feb 23, 2014 at 23:46
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2$\begingroup$ Nice illustration, but how does it answer the question? $\endgroup$ Commented Feb 25, 2014 at 18:59
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$\begingroup$ This isn't rigorous, you haven't defined "low" and "high" and the scale of the gradient! $\endgroup$– ThomasCommented Feb 26, 2014 at 11:34
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1$\begingroup$ @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. $\endgroup$ Commented Feb 26, 2014 at 12:47
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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
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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.
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1$\begingroup$ … 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. $\endgroup$ Commented Oct 24, 2014 at 17:41
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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.
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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.
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Firstly for a point mass: mass x radius squared is a measure of moment of inertia, and secondly all the point masses moment of inertia can always be added up to give the total effective moment of inertia.
Now the two sphere have same radius, but different mass distribution. If all the mass is concentrated along the radius, the angular momentum measure is definitely going to be larger than the angular momentum of any other points with smaller radius.
