How can I find the moment of inertia for a simple shape?

[Zorodius]

I know the moment of inertia is given by:

$$I = \int r^2~dm$$

I don't understand how I can apply this, though. Could someone give me an example or a description of the process I would go through to find the moment of inertia for any simple shape - a solid cylinder or whatever? What do I need to know about an object before I can find its moment of inertia?

 

[Doc Al]

You need to know how the mass is distributed with respect to the axis of rotation. That mass element "dm" needs to be written in terms of "dr". If the shape is simple, and the density is uniform, then the integral may be easy to calculate. You can look up formulas for the moments of inertia of standard shapes, like disks, balls, and rods.

Here's an example of how to calculate the moment of inertia of a thin rod rotating around its center of mass: http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod3

 

[Zorodius]

Thanks for the reply!

I think I have a somewhat better idea now, but I still don't understand how to evaluate the integral, even for a simple object with uniform density.

Do I need to know multivariable calculus to do this?

What's wrong with the following? I realize it's not correct, because the answer is wrong, but I don't know why it doesn't work:

Suppose I have a cylinder of uniform density rotating about an axis parallel to the cylinder and passing through the center of the cylinder. Then, since the density is uniform, if M = total mass and V = total volume,

$$\frac{dm}{dV} = \frac {M}{V}$$

$$dm = M \frac {dV}{V}$$

For a cylinder with radius r and height h,

$$V = \pi h r^2$$

$$\frac{dV}{dr} = 2 \pi h r$$

$$dV = 2 \pi h r~dr$$

Substituting this into the earlier equation, and substituting the formula for volume in place of V,

$$dm = M\frac {2 \pi h r ~dr}{\pi h r^2} = M\frac {2~dr}{r}$$

Substituting that into the formula for the moment of inertia and integrating from 0 to R, where R is the radius of the entire cylinder (... right?),

$$I = \int_0^R r^2~dm = \int_0^R r^2~M\frac {2~dr}{r} = 2M \int_0^R r~dr$$

$$I = 2M \left[ \frac{r^2}{2} \right]_0^R = MR^2$$

... which is, of course, wrong. :yuck:

 

[e(ho0n3]

Suppose I have a cylinder of uniform density rotating about an axis parallel to the cylinder and passing through the center of the cylinder. Then, since the density is uniform, if M = total mass and V = total volume,

$$\frac{dm}{dV} = \frac {M}{V}$$

$$dm = M \frac {dV}{V}$$

For a cylinder with radius r and height h,

$$V = \pi h r^2$$

$$\frac{dV}{dr} = 2 \pi h r$$

$$dV = 2 \pi h r~dr$$

Substituting this into the earlier equation, and substituting the formula for volume in place of V,

$$dm = M\frac {2 \pi h r ~dr}{\pi h r^2} = M\frac {2~dr}{r}$$

You made a slight error here in the last equation. Note that $V = \pi R^2h$ so

$$dm = M\frac {2 \pi h r ~dr}{\pi h R^2} = \frac{M}{R^2} 2r~dr$$

and the rest follows.

 

[Zorodius]

Thanks a lot for your reply, e(ho0n3! I really appreciate it!

Not to be a pain in the neck, but could someone tell me why the following doesn't work either?

Suppose a solid sphere with uniform density is rotating around an axis passing through the center of the sphere. Then,

$$V = \frac {4}{3} \pi r^3$$

$$dV = 4 \pi r^2 ~ dr$$

Since the sphere is of uniform density, then:

$$\frac {dm}{dV} = \frac {M}{V}$$

$$dm = M \frac {dV}{V} = M \frac {4 \pi r^2 ~ dr}{\frac{4}{3} \pi R^3} = \frac {3M}{R^3} \cdot r^2~dr$$

Where R is the radius of the sphere. Then, using the definition of the moment of inertia,

$$I = \int r^2 dm = \frac {3M}{R^3} \int_0^R r^4~dr = \frac {3M}{R^3} \left[ \frac {r^5}{5} \right]_0^R = \frac {3}{5} MR^2$$

It should be 2/5, not 3/5. Why doesn't this work? Did I make (another) error in calculation, or is there some other reason why this isn't a valid way of finding the moment of inertia of a sphere?

 

[DarkEternal]

make sure you are considering the fact that r is measured from the axis of rotation, not the center. i think this might be your problem.

$$a = r \sin \phi$$
$$\rho = \frac{M}{\frac{4}{3} \pi R^3}$$
$$I=\int a^2 dm=\iiint a^2 \rho dV=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R} a^2 \rho r^2 \sin \phi dr d\phi d\theta$$
$$=\frac{2}{5} M R^2$$

ehhhh forgive the messy latex, I've never used it before

 

[Doc Al]

It should be 2/5, not 3/5. Why doesn't this work? Did I make (another) error in calculation, or is there some other reason why this isn't a valid way of finding the moment of inertia of a sphere?

In the definition of rotational inertia, r refers to the distance from the axis of rotation. But in your integral, you are using r to refer to the distance from the center of the sphere. You'll need to rethink your volume element in terms of distance from the axis.

 

[Zorodius]

Is it correct to say that the process by which you arrive at the moment of inertia for a sphere and many other objects is to first break the object down into volumes whose moments of inertia are known (like cylinders or cylindrical discs), and then add up all those moments of inertia to find the moment of inertia of your entire object?

I'm trying to follow the way they arrive at the moment of inertia for a sphere on HyperPhysics. They seem to be adding up the moments of inertia for a bunch of discs stacked on top of one another (like in the sphere on the left in the attachment). Would it also be correct to instead add up the moments of inertia for a bunch of cylinders wrapped around the axis of rotation? (like the sphere in the right in the attachment.)

 

[DarkEternal]

Is it correct to say that the process by which you arrive at the moment of inertia for a sphere and many other objects is to first break the object down into volumes whose moments of inertia are known (like cylinders or cylindrical discs), and then add up all those moments of inertia to find the moment of inertia of your entire object?

I'm trying to follow the way they arrive at the moment of inertia for a sphere on HyperPhysics. They seem to be adding up the moments of inertia for a bunch of discs stacked on top of one another (like in the sphere on the left in the attachment). Would it also be correct to instead add up the moments of inertia for a bunch of cylinders wrapped around the axis of rotation? (like the sphere in the right in the attachment.)

if you look at the formula, $$I=\int r^2dm$$, it's easy to see that basically you are adding up all the moments of inertia of all the point masses that comprise the whole mass (for a point mass, $$I=mr^2$$). so as long as whatever method you use to break down the mass (discs, cylinders, etc.) accounts for every point's moment of inertia, then it should come out with the correct answer. i find it is easiest to integrate the formula for distance to the axis of rotation squared times the density over the volume of the object.

 

[Zorodius]

I appreciate the replies, but I've been trying to understand this for hours and hours and it's still clear as mud. I can (barely) follow along with the way they find moments of inertia on HyperPhysics, but I can't figure out how to do it on my own. I think I'm running into problems because my understanding of calculus beyond derivatives is so poor. Could someone please suggest what I should study to understand this better - regular integral calculus? Multivariable calculus? Differential equations?

And if you're aware of any good tutorials on the web for whatever subjects I should study to understand this better, I would appreciate a link.

 

[HallsofIvy]

Since "moment of intertia" involves solids, multivariable calculus is the correct place to look.

As Doc Al pointed out- the "r" in calculating moment of inertia is distance from the axis while the "r" in the sphere is distance from (0,0,0). It is almost always best to set the problem in cylindrical coordinates with the z-axis along the axis of rotation.

In cylindrical coordinates, x²+ y²+ z²= R² becomes r²+ z²2= R².