# Homework Help: Prove that a cylinder has the least moment of inertia

1. Nov 24, 2012

### whyme1010

1. The problem statement, all variables and given/known data
Consider a solid homogeneous object taller than its wide, with the top and bottom parallel. Prove that the shape with the lowest moment of inertia about an axis through the object is a cylinder with the axis through its axis of symmetry.

density, mass, and height are constant.

2. Relevant equations
∫p^2dm = I, where p is the radius and dm is the differential of mass.

3. The attempt at a solution
Maybe start with a cylinder and show that decreasing the radius of any disc of height dh and increasing the radius of another disc of height dh results in an increase in the moment of inertia?
The problem with this question is that it doesn't say anything about how symmetric the shape is.

2. Nov 24, 2012

### Staff: Mentor

A cylinder covers all the volume with a distance d<R of the axis and nothing else.
If you have an arbitrary object which has some point outside that radius, and the volume is constant, you have a "hole" closer to the axis. How can you use this to decrease the moment of inertia?

3. Nov 24, 2012

### rcgldr

If the solid object is a solid homogeneous of constant mass and density, the top and bottom are parallel, and height is constant, doesn't this restrict the shape to a cylinder?

If the entire outer edge surface across the width of the object do not have to be parallel, then then two "wide" and "short" cones connected base to base would have less inertia. The two cone bases could be attached to a very thin cylinder in order to have the "top" and "bottom" parallel.

Last edited: Nov 24, 2012
4. Nov 24, 2012

### whyme1010

well a double hemisphere would have the least moment of inertia.

so I'm pretty sure it's asking for a uniform cross section body.

But then this entire question is trivial.

5. Nov 24, 2012

### haruspex

No, a rectangular block would satisfy those conditions.
It would not have the least within the bounds stated.
I would tackle this in stages. First, try to show that the horizontal cross section is always a circle. Then show the circle must have the same radius at all heights.

6. Nov 24, 2012

### whyme1010

Well to prove that each cross section disc has to be a circle, I would start by assuming that it was a circle to start with. Now take any infinitesimal amount of mass on the circumference and move it to a radius greater than that of the circle. by the way moment of inertia is defined, this would increase the total moment of inertia.

An argument for the uniformity of the cylinder would be to start with one disc of height H, density D, and mass M.
Assume a constant height, density, and mass.
This one disc would be a cylinder.
Now look at two discs. Then M = piR^2(h) + piR^2(H-h)
Now increase the radius of one disc by radius r. It can be shown that the moment of inertia increases.
This is also a cylinder.
Now add a third disc and so on.
By induction, the shape has to be a cylinder.

Last edited: Nov 24, 2012
7. Nov 24, 2012

### whyme1010

Not sure if this is correct. Or even how to write it up.

8. Nov 24, 2012

### haruspex

Yes, that's the sort of thing. Can't use induction as such because you're not dealing with a countable infinity of possibilities, but you can use reductio ad absurdum: if the radius is different for thin discs at two heights, show that evening them out (preserving total volume) reduces MI.
Similarly, I would word the first part around the other way: suppose it is not a circle and show you can reduce the MI.

9. Nov 25, 2012

### rcgldr

The "height" of a rectangle or any shape where the cross section is not circular would vary at it rotates.

10. Nov 25, 2012

### tiny-tim

mfb at post #2 has already shown the correct way!

11. Nov 25, 2012

### haruspex

I admit it could be clearer, but I should think the question intends a vertical axis.

12. Nov 25, 2012

### haruspex

I had not been able to read that post, but now I discover the trick is to mouse over it. (What's the purpose of that format?)

13. Nov 25, 2012

### tiny-tim

to give a two-stage hint

saves waiting for a second post (which could be overnight) if the first-stage hint isn't enough!