Rotational Inertia

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

Discussion

introduction & theory

Logic behind the moment of inertia: Why do we need this?

	translational		connections		rotational
Translational and Rotational Quantities Compared
cause of acceleration	∑ F		τ =	r × F	∑ τ
resistance to acceleration	m		I =	∑ r_i²m_i = ∫ r² dm	I
newton's second law	∑ F =	m a			∑ τ =	I α

Definition for point bodies

I = mr²

It's a scalar quantity (like its translational cousin, mass), but has unusual looking units.

[kg m²]

Say it, kilogram meter squared and don't say it some other way by accident.

For a collection of objects, just add the moments. It works like mass in this respect as long as you're adding moments that are measured about the same axis.

I = ∑I = ∑mr²

For an extended body, replace the summation with an integral and the mass with an infinitesimal mass. You add up (integrate) all the moments of inertia contributed by the teeny, tiny masses (dm) located at whatever distance (r) from the axis they happen to lie.

I =

⌠
⌡

r² dm

In practice, for objects with uniform density (ρ = m/V) you do something like this …

I =	⌠ ⌡	r² dm =	⌠ ⌡	r² ρ dV =	⌠ ⌡	r²	m	dV
							V

For objects with nonuniform density, replace density with a density function, ρ(r).

I =

⌠
⌡

r² dm =

⌠
⌡

r² ρ(r) dV

The infinitesimal quantity dV is a teeny tiny piece of the whole body. In practice, this may take one of two forms (but it is not limited to these two forms). The infinitesimal box is probably the easiest conceptually. Imagine dicing the object up into cubes.

[photo of cubed potatoes]

The pieces are dx wide, dy high, and dz deep. The volume of each infinitesimal piece is …

dV = dx dy dz

When an object is essentially rectangular, you get a set up something like this …

I =	⌠⌠⌠ ⌡⌡⌡	(x² + y² + z²)	m	dx dy dz	or	I =	⌠⌠⌠ ⌡⌡⌡	(x² + y² + z²) ρ(x, y, z) dx dy dz
			V

This is the way to find the moment of inertia for cubes, boxes, plates, tiles, rods and other rectangular stuff. Note that although the strict mathematical description requires a triple integral, for many simple shapes the actual number of integrals worked out through brute force analysis may be less. Sometimes, the integrals are trivial.

The other easy volume element to work with is the infinitesimal tube. Imagine a leek.

[photo of a leek]

Each layer of the leek has a circumference 2πr, thickness dr, and height h. The volume of each infinitesimal layer is then …

dV = 2πrh dr

For many cylindrical objects, you basically start with something like this …

I =	⌠ ⌡	r²	m	2πrh dr	or	I =	⌠ ⌡	r² ρ(r) 2πrh dr
			V

This method can be applied to disks, pipes, tubes, cylinders, pencils, paper rolls and maybe even tree branches, vases, and actual leeks (if they have a simple mathematical description).

When shapes get more complicated, but are still somewhat simple geometrically, break them up into pieces that resemble shapes that have already been worked on and add up these known moments of inertia to get the total.

I_total = I₁ + I₂ + I₃ + …

For slightly more complicated round shapes, you may have to revert to an integral that I'm not sure how to write. Something like for nested, cylindrical shells …

I =

⌠
⌡

I_{cylindrical shell}(r) dr

or this for stacked disks and washers

I =

⌠
⌡

I_{disk or washer}(r) dr

These methods can be used to find the moment of inertia of things like spheres, hollow spheres, thin spherical shells and other more exotic shapes like cones, buckets, and eggs -- basically, anything that might roll and that has a fairly simple mathematical description.

When you are done with all of this, you oftentimes end up with a nice little formula that looks something like this …

I = αmr²

where α is a simple rational number like 1 for a hoop, ½ for a cylinder, or ⅖ for a sphere.

What if an object isn't being rotated about the axis used to calculate the moment of inertia? Apply the parallel axis theorem.

I = I_cm + mL²

What can I say about the perpendicular axis theorem other than it's interesting, but I've never used it? Maybe, that it works for laminar objects only.

I_z = I_x + I_y

The best way to learn how to do this is by example. Lots of examples.

Summary

bullet

Problems

practice

Four point objects of mass m are located at the corners of a square of side s as shown in the figure to the right. Determine the moment of inertia of this system if it is rotated about …

the perpendicular bisector of a side
a side
a diagonal
one corner on an axis perpendicular to the plane containing the masses

Solutions …

In the first case, each of the four masses is a distance s/2 from the axis. Thus …


I = ∑ r²m = 4	⎛ ⎝	s√2	⎞² ⎠	m = ms²
I = ∑ r²m = 4	⎛ ⎝	2	⎞² ⎠	m = ms²

In the second case, two of the masses are on the axis and contribute nothing to the moment of inertia. The other masses are each s away from the axis. Thus …

I = ∑ r²m = 2ms²

In the third case, two masses lie on the axis and two are half a diagonal away from the axis. Thus …


I = ∑ r²m = 2	⎛ ⎝	s√2	⎞² ⎠	m = ms²
I = ∑ r²m = 2	⎛ ⎝	2	⎞² ⎠	m = ms²

In the fourth case, one mass lies on the axis, two masses are a distance s away, and one is a diagonal away from the axis. Thus …

I = ∑ r²m = 2ms² + m(s√2)² = 4ms²

Write something.
- Answer it.
Write something.
- Answer it.

Determine the moment of inertia for each of the following shapes. The rotational axis is the same as the axis of symmetry in all but two cases. Use M for the mass of each object.

ring, hoop, cylindrical shell, thin pipe
annulus, hollow cylinder, thick pipe
disk, solid cylinder
spherical shell
hollow sphere
solid sphere
rod, rectangular plate (perpendicular bisector)
rod, rectangular plate (axis along edge)
rectangular plate, solid box (axis perpendicular to face)
cube (axis perpendicular to face)
cone (rotated about its central axis)
cone (rotated about its vertex)

[magnify]

ring, hoop, cylindrical shell, thin pipe

There isn't much of a proof here. Since all the mass is located the same distance R away from the axis of rotation, the moment of inertia is the same as that for a point mass located a distance R from the axis, namely …

I = ⌠
⌡ r² dm = R² ⌠
⌡ dm

which has a trivial solution …

I = MR²

Note how the height of the hoop is not a factor. This formula would work equally well for a long thin tube or a flat thin ring.
annulus, hollow cylinder, thick pipe

A hollow cylinder is basically a series of infinitesimally thin nested cylindrical shells all added together. The way to write this in calculus is …

I = ⌠
⌡ r² dm

The mass of each infinitesimal slice (dm) is the overall density (ρ) times the infinitesimal volume (dV) of the slice.

I = ⌠
⌡ r² ρ dV

The infinitesimal volume is the surface area of a cylindrical shell (2πrh) times its infinitesimal thickness (dr).

I = ⌠
⌡ r² ρ 2πrh dr

The last piece of the puzzle is density, which is mass divided by volume.

I = ⌠
⌡ r² M 2πrh dr

V

The volume of a hollow cylinder is the volume of the outer cylinder minus the volume of the inner cylinder.

V = πR₂²h − πR₁²h = π (R₂² − R₁²) h

Putting it altogether and integrating from the inner radius (R₁) to the outer radius (R₂) yields …

R₂

I = ⌠
⌡ r² M 2πrh dr = 2M ⌠
⌡ r³ dr = 2M R₂⁴ − R₁⁴

π (R₂² − R₁²) h R₂² − R₁² R₂² − R₁² 4

R₁

which simplifies to …

I = M (R₂² + R₁²) (R₂² − R₁²)

2 (R₂² − R₁²)

and eventually simplifies to …

I = 1 M (R₂² + R₁²)

2

Note how height cancelled out of this equation a few steps back. This formula would work for a long, thick-walled pipe or a flat, hollowed out disk (also known as an annulus).
disk, solid cylinder

A solid cylinder is a hollow cylinder with an inner radius of zero, so this proof is similar to the previous one. Start with the definition of the moment of inertia and substitute density times volume (ρ dV) for mass (dm).

I = ⌠
⌡ r² dm = ⌠
⌡ r² ρ dV

The infinitesimal volume is the surface area of a cylindrical shell (2πrh) times its infinitesimal thickness (dr). The density of a uniform cylinder is its total mass (M) divided by its total volume (πR²h).

I = ⌠
⌡ r² ρ dV = ⌠
⌡ r² M 2πrh dr

πR²h

Now, integrate all the infinitesimal shells from r = 0 to r = R …

R

I = 2M ⌠
⌡ r³ dr = 2M R⁴

R² R² 4

0

and simplify …

I = 1 MR²

2

Once again, height is not a factor affecting the moment of inertia of this shape. This formula would work for a long solid cylinder or a flat solid disk.
spherical shell

This is a tough proof. As always, start with the basic formula.

I = ⌠
⌡ r² dm = ⌠
⌡ r² ρ dV

Now the hard part. How do we slice this thing up? I recommend rings. Imagine the standard unit circle from trig class. Start on the x axis as is the usual way and walk counterclockwise across the circumference of the circle measuring and angle θ that starts at 0 radians and ends at π radians taking teeny, tiny dθ steps. (I'll use the x axis as the axis of rotation. I hope that's OK.) The radius of each ring is R sin θ, which means its circumference is 2πR sin θ. The width of one of these rings would be R dθ and its thickness would be something small. Something that will hopefully go away in the math we're about to start. Let's call it t. This gives us a volume element dV = (2πR sin θ)(R dθ)(t) and an integral …

I = ⌠
⌡ (R sin θ)² ρ (2πR sin θ Rdθ t)

We're getting closer. Replace density with mass per volume. The volume of a spherical shell would equal the surface area of the shell (4πR²) times its thickness (t).

I = ⌠
⌡ (R sin θ)² m (2πR sin θ R dθ t) = ⌠
⌡ (R sin θ)² M (2πR sin θ R dθ)

V 4πR² t

Clean this beast. I beg you.

I = MR² ⌠
⌡ sin³ θ dθ

2

Wow! What happened to all the symbols? I'm telling you this algebra stuff is magic. Oops, I forgot the limits of integration. Let's put them in.

π

I = MR² ⌠
⌡ sin³ θ dθ

2

0

Hmmm, I don't quite know how to solve this one. May I suggest looking up the result in an integral table? Or maybe, perhaps, letting a machine do the work for you? If you tell this one to find the integral of (Sin[x])^3 it will return something like this expression without the constants in the front or the limits at the end …

π

I = MR² 1 ⎡
⎣ cos 3θ − 9 cos θ ⎤
⎦

2 12

0

The limits of this integral are … well … something. I feel so lazy today after finding all these moments of inertia. Let me use another online source to calculate the upper limit …

cos(3 * pi) - (9 * cos(pi)) = 8

and the lower limit …

cos(3 * 0) - (9 * cos(0)) = -8

of the quantity in the square bracket.

I = MR² 1 ⎡
⎣ (+8) − (−8) ⎤
⎦ = 16 MR²

2 12 24

I see the final answer approaching.

I = 2 MR²

3

I am now officially happy.
hollow sphere

What is a hollow sphere but a series of spherical shells piled on top of one another. Do not use the basic formula.

do not use I = ⌠
⌡ r² dm do not use

Start with something we just dervied a second ago -- the moment of inertia of a spherical shell.

I_{spherical shell} = 2 MR²

3

Break the hollow sphere up into a series of infinitesimal spherical shells and integrate these infinitesimal moments.

R₂

I = ⌠
⌡ 2 r² dm

3

R₁

Replace dm with ρ dV. Replace density with total mass (M) over total volume (⁴/₃ π (R₂³ − R₁³)). Replace dV with the surface are of a sphere (4πr²) times its infinitesimal thickness (dr).

R₂ R₂

I = ⌠
⌡ 2 r² ρ dV = ⌠
⌡ 2 r² M 4πr² dr

3 3 ⁴/₃ π (R₂³ − R₁³)

R₁ R₁

This can be simplified to …

R₂

I = 2M ⌠
⌡ r⁴ dr

R₂³ − R₁³

R₁

which certainly is simple to integrate.

R₂

I = 2M ⎡
⎣ r⁵ ⎤
⎦

R₂³ − R₁³ 5

R₁

Put the limits in …

I = 2M R₂⁵ − R₁⁵

R₂³ − R₁³ 5

and clean it up a bit.

I = 2 m R₂⁵ − R₁⁵

5 R₂³ − R₁³

This is as simple as I can make it.

solid sphere

You want an easy proof? What is a solid sphere but a hollow sphere with no inner radius. Start with the hollow sphere formula

I_{hollow sphere} =	2	m	R₂⁵ − R₁⁵
	5		R₂³ − R₁³

Let R₂ = R and take the limit as R₁ → 0

I =	2	m	R⁵
	5		R³

Simplify and we're done.

You want a harder proof? A solid sphere is built like an onion from layer upon layer of thin spherical shells. Each shell has moment of inertia equal to

I_{spherical shell} =	⌠ ⌡		2	r²dm =	⌠ ⌡		2	r² ρ dV
			3				3

Again, density is total mass (M) divided by total volume (⁴/₃ πR³) and infinitesimal volume (dV) is the surface area of a spherical shell (4πr²) times its infinitesimal thickness (dr). Substitute these values and simplify …

I =	⌠ ⌡		2	r²	M	4πr² dr = 2M	⌠ ⌡	r⁴ dr
			3		⁴/₃ πR³

Yet another simple integral …

	R				R
I = 2M	⌠ ⌡	r⁴ dr = 2M	⎡ ⎣	r⁵	⎤ ⎦
I = 2M	⌠ ⌡	r⁴ dr = 2M	⎡ ⎣	5	⎤ ⎦
	0				0

and it gives us the right answer …

Dare I try another proof? What is a solid sphere but a stack of disks.

I_disk =	⌠ ⌡		1	r² dm =	⌠ ⌡		1	r² ρ dV
			2				2

Review your analytical geometry. The formula for a circle is …

R² = x² + y²

The disks of our sphere have radii (represented by the symbol y) that vary according to this formula.

y² = R² − x²

Again, density is total mass (M) divided by total volume (⁴/₃ πR³), but now the infinitesimal volume (dV) is the surface area of a circular disk (πy²) times its infinitesimal thickness (dx). Substitute, simplify, …

I =	⌠ ⌡		1	(R² − x²)	M	π(R² − x²) dx =	3M	⌠ ⌡	(R² − x²)² dx
			2		⁴/₃ πR³		8R

and integrate. It's an ugly one. Viewer discretion is advised.

I =

⌠
⌡

(R² − x²)² dx =

⎡
⎣

x⁵

−

2R²x³

+ R⁴x

⎤
⎦

16R⁵

8R³

−R

All of the stuff in square brackets reduces to 16/15 R⁵. Trust me. I've checked it several times. One last bit of simplification and we're done.

rod, rectangular plate (perpendicular bisector)

Let M and L be the mass and length of the plate respectively. Then …

λ = M

L

is its linear density. Divide the rectangle up into thin strips that run parallel to the axis of rotation. The width of these strips, dx, times the linear density is the infinitessimal mass of each. Plop this into the moment of inertia formula and integrate from the left edge of the plate (−½L) to the right edge (+½L).

+½L +½L

I = ⌠
⌡ r² dm = ⌠
⌡ x² λ dx = ⌠
⌡ x² M dx = ⎡
⎣ Mx³ ⎤
⎦

L 3L

−½L −½L

Stuff cancels, and with a minimal amount of work you end with …

I = 1 ML²

12
rod, rectangular plate (axis along edge)

Use the same set up as in the previous proof. Integrate from the left edge of the plate to the right edge; that is, from 0 to L.

L L

I = ⌠
⌡ r² dm = ⌠
⌡ x² λ dx = ⌠
⌡ x² M dx = ⎡
⎣ Mx³ ⎤
⎦

L 3L

0 0

Easy peasy, here's the answer …

I = 1 ML²

3

You could also try using the parallel axis theorem.

I = I_cm + mL²

The moment of inertia about the center of mass was determined in the previous proof. Just add on a little correction and we're done.

I = 1 ML² + m(½L)² = ⎛
⎝ 1 + 1 ⎞
⎠ ML²

12 12 4

This simplifies to the answer …

I = 1 ML²

3

rectangular plate, solid box (axis perpendicular to face)

Start with the basic formula, but make one sup change. We'll replace the volume density (ρ = M/V) with surface density (σ = M/A) since the thickness of the plate doesn't contribute anything to the moment of inertia about this axis.

I =	⌠ ⌡	r² dm =	⌠ ⌡	r² σ dA =	⌠ ⌡	r²	M	dA
							A

Now let's dice the plate up into rectangular strips dx long by dy wide and any old height whatsoever.

I =	⌠⌠ ⌡⌡	r²	M	dx dy
			LW

Since I like food preparation analogies, imagine we're slicing the plate up into infinitessimal french fries. Each french fry has coordinates (x, y) relative to the axis, which means their distances from the axis can be found using pythagoras' theorem.

r² = x² + y²

Now, put everything altogether and set the limits of integration. For a plate of length L and width W, the appropriate limits would be ±½L and ±½W.

	_+½W	_+½L
I =	⌠ ⌡	⌠ ⌡	(x² + y²)	M	dx dy
I =	⌠ ⌡	⌠ ⌡	(x² + y²)	LW	dx dy
	^−½W	^−½L

Integrate first over x while y stays constant …

		_+½W					_+½W
I =	M	⌠ ⌡	x³	+ xy²	dy =	M	⌠ ⌡	L³	+ Ly²	dy
I =	LW	⌠ ⌡	3	+ xy²	dy =	LW	⌠ ⌡	12	+ Ly²	dy
		^−½W					^−½W

then integrate over y …

_+½W

I =

⎡
⎢
⎣

L³y

Ly³

⎤
⎥
⎦

⎛
⎜
⎝

L³W

LW³

⎞
⎟
⎠

^−½W

and then simplify.

  
     I = 
    1
     M (L2 + W2)
  
    12

cube (axis perpendicular to face)

A cube is a plate with length and width equal. Start with the results of the previous proof …

I = 1 M (L² + W²)

12

and set L = W = S.

I = 1 M (S² + S²)

12

Voila, c'est finis!

I = 1 MS²

6
cone (rotated about its central axis)

A cone is a series of infinitesimally thin disks of varying radius. If we add up the moments of inertia of all these very, very thin slices we'll get the moment of inertia of the whole cone. Adding up a lot of very small pieces to create a whole is called integration.

I = ⌠
⌡ I_slice dx = ⌠
⌡ 1 m_slicer² dx

2

Replace mass with density times volume and proceed.

I = ⌠
⌡ 1 ρA r² dx = ⌠
⌡ 1 ρ (πr²) r² dx = ⌠
⌡ 1 πρr⁴ dx

2 2 2

The "trick" to solving this part of the problem is determining how the radius of the slices vary from the vertex (x = 0) to the base (x = H). We need a function that begins at 0, ends at R, and increases linearly. May I suggest …

r = R x

H

Make the switch and integrate.

H H

I = ⌠
⌡ 1 πρ ⎛
⎝ R x ⎞⁴
⎠ dx = πρR⁴ ⎡
⎣ x⁵ ⎤
⎦ = πρR⁴H

2 H 2H⁴ 5 10

0 0

Recall that the volume of a cone is …

V = 1 πR²H

3

Do you see the volume hidden inside the moment of inertia? It's in there.

I = πρR⁴H = ρ ⎛
⎝ 1 πR² H ⎞
⎠ ⎛
⎝ 3 R² ⎞
⎠ = ρV 3 R²

10 3 10 10

Density times volume is mass. Therefore …

I = 3 MR²

10
cone (rotated about its vertex)

Here's the answer …

I = 3 M ⎛
⎝ 1 R² + H² ⎞
⎠

5 4

I'll leave it to the bold reader to work out the solution. Gotta leave something for homework.

conceptual

Spacecraft from the US landed on Mars in 1976 and 1997. By communicating with spacecraft on the surface, NASA scientists were able to determine the orientation of Mars in space. Changes in the orientation between 1976 and 1997 were then used to determine the moment of inertia of Mars. Why would anyone care about the moment of inertia of a planet? What good is it to know this quantity?

numerical

Write something.
Write something else.
Write something completely different.

algebraic

For many applications, it's better to have the moment of inertia written in terms of the density of the material it's made out of instead of the mass of the finished object. Do this for each of the following shapes …

object

mass-moment

density-moment

annulus, hollow cylinder, thick pipe

I =	1	M (R₂² + R₁²)
	2

disk, solid cylinder

I =	1	MR²
	2

hollow sphere

I =	2	m	R₂⁵ − R₁⁵
	5		R₂³ − R₁³

solid sphere

I =	2	MR²
	5

rectangular plate, solid box (axis perpendicular to face)

I =	1	M (L² + W²)
	12

cube (axis perpendicular to face)

I =	1	MS²
	6

cone (rotated about its central axis)

I =	3	MR²
	10

cone (rotated about its vertex)

I =	3	M	⎛ ⎝		1	R² + H²	⎞ ⎠
	5				4

calculus

Show that the moment of inertia of a cone rotated about its vertex is given by …

I =	3	M	⎛ ⎝		1	R² + H²	⎞ ⎠
	5				4

statistical

The data in the text file earth.txt gives the density and gravitational field strength of the earth at various depths below the surface. Using data analysis software (preferably something that can do numerical integration) determine the moment of inertia of the earth.

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Glenn Elert

						R₂
I =	⌠ ⌡	r²	M	2πrh dr =	2M	⌠ ⌡	r³ dr =	2M	R₂⁴ − R₁⁴
I =	⌠ ⌡	r²	π (R₂² − R₁²) h	2πrh dr =	R₂² − R₁²	⌠ ⌡	r³ dr =	R₂² − R₁²	4
						R₁

I =	M (R₂² + R₁²) (R₂² − R₁²)
	2 (R₂² − R₁²)

I =	⌠ ⌡	(R sin θ)²	m	(2πR sin θ R dθ t) =	⌠ ⌡	(R sin θ)²	M	(2πR sin θ R dθ)
			V				4πR² t

					π
I =	MR²	1	⎡ ⎣	cos 3θ − 9 cos θ	⎤ ⎦
I =	2	12	⎡ ⎣	cos 3θ − 9 cos θ	⎤ ⎦
					0

					+½L						+½L
I =	⌠ ⌡	r² dm =	⌠ ⌡	x² λ dx =	⌠ ⌡	x²	M	dx =	⎡ ⎣	Mx³	⎤ ⎦
I =	⌠ ⌡	r² dm =	⌠ ⌡	x² λ dx =	⌠ ⌡	x²	L	dx =	⎡ ⎣	3L	⎤ ⎦
					−½L						−½L

I =	⌠ ⌡	1	ρA r² dx =	⌠ ⌡	1	ρ (πr²) r² dx =	⌠ ⌡	1	πρr⁴ dx
		2			2			2