Moment of Inertia of Discrete and Continuous Bodies

Moment of Inertia of Discrete and Continuous Bodies

For translational motion, mass measures resistance to acceleration.

For rotation about a specified axis, the corresponding mass-distribution quantity is the moment of inertia.

For a collection of particles rotating about a fixed axis,

(1)

where r_i⊥ is the perpendicular distance from particle i to the rotation axis.

For a continuous body, the sum becomes an integral:

(2)

The square on the distance is crucial. Moving mass farther from the axis increases the moment of inertia strongly.

Moment of inertia therefore depends on:

• total mass,
• shape,
• distribution of mass,
• location and direction of the chosen axis.

It is not a property of a body alone. It is a property of a body about a specified axis.

1 Why the distance is squared

Consider a point mass m rotating with angular speed ω at perpendicular radius r_⊥.

Its speed is

(3)

Its kinetic energy is

K	= mv2	(4)
	= m 2	(5)
	= mr⊥2ω2.	(6)

For many particles in the same rigid body rotation,

K	= ∑ imiri⊥2ω2	(7)
	= ω2.	(8)

This motivates the definition

(9)

so that

(10)

The squared distance is therefore a direct consequence of the relation

(11)

2 Axial angular momentum connection

For a particle rotating about the z axis,

(12)

Summing over all particles,

(13)

Thus

(14)

The same mass-distribution quantity therefore appears in both rotational kinetic energy and axial angular momentum.

3 Moment of inertia depends on the axis

Suppose the same body is rotated about two different axes.

The particle distances r_i⊥ generally change.

Therefore

(15)

in general.

Even if the axis direction is unchanged, translating the axis to a different location usually changes I.

This axis dependence is developed systematically in M05-05 using the parallel-axis theorem.

4 Units and dimensions

From

(16)

the SI unit is

(17)

The dimensions are

(18)

Moment of inertia is not a torque, energy, or angular momentum.

5 Mass near the axis versus mass far from the axis

Because

(19)

doubling a particle’s radius multiplies its contribution to I by four.

Tripling the radius multiplies the contribution by nine.

For equal masses:

(20)

A relatively small amount of mass far from the axis can contribute more rotational inertia than a larger amount of mass close to the axis.

6 Example 1: three point masses

Three point masses rotate about the z axis:

m1	= 2 kg,	r1	= 0.50 m,	(21)
m2	= 3 kg,	r2	= 1.00 m,	(22)
m3	= 1 kg,	r3	= 2.00 m.	(23)

Then

Iz	= m1r12 + m 2r22 + m 3r32	(24)
	= 2(0.50)2 + 3(1.00)2 + 1(2.00)2	(25)
	= 0.50 + 3.00 + 4.00.	(26)

Therefore

(27)

Although m₃ is the smallest mass, it contributes

(28)

more than either of the other particles, because it is farthest from the axis.

7 From a discrete sum to a continuous integral

A continuous body can be imagined as divided into many small mass elements

(29)

The approximate moment of inertia is

(30)

Take the limit as each element becomes infinitesimal:

(31)

This is the fundamental definition for continuous mass distributions.

8 Choosing the correct mass element

To evaluate

(32)

the mass element dm must be expressed in terms of the chosen geometric coordinates.

The appropriate density depends on the dimensional character of the idealized body.

8.1 Linear mass density

For a thin wire or rod,

(33)

Therefore

(34)

For uniform linear density,

(35)

8.2 Surface mass density

For a thin plate or lamina,

(36)

Therefore

(37)

For uniform surface density,

(38)

8.3 Volume mass density

For a three-dimensional body,

(39)

Therefore

(40)

For uniform volume density,

(41)

9 General integration workflow

A reliable moment-of-inertia calculation follows the sequence:

1. Specify the rotation axis.
2. Choose coordinates adapted to the geometry.
3. Write r_⊥, the perpendicular distance from the mass element to the axis.
4. Express dm using λ, σ, or ρ.
5. Set the integration limits.
6. Evaluate

   (42)

7. Check units and limiting behavior.

The geometry of r_⊥ is often the most important step.

10 Uniform thin rod about its center

Consider a thin rod of length L and mass M lying along the x axis from

(43)

to

(44)

Let the rotation axis pass through the rod’s center and be perpendicular to the rod.

The linear density is

(45)

A small element has mass

(46)

Its perpendicular distance to the axis is

(47)

Therefore

I	= ∫ −L∕2L∕2x2 dm	(48)
	= ∫ −L∕2L∕2x2 dx	(49)
	= −L∕2L∕2.	(50)

Thus

(51)

11 Uniform thin rod about one end

Now place the perpendicular rotation axis through one end of the same rod.

Choose

(52)

Then

Iend	= ∫ 0Lx2 dx	(53)
	= .	(54)

Therefore

(55)

The same rod has a larger moment of inertia about its end because more of its mass lies farther from the axis.

In M05-05 this result will be recovered immediately using the parallel-axis theorem.

12 Thin ring about its symmetry axis

Consider a thin ring of radius R and total mass M.

Every mass element lies at the same perpendicular distance:

(56)

Therefore

I	= ∫ R2 dm	(57)
	= R2 ∫ dm	(58)
	= MR2.	(59)

Thus

(60)

No detailed integration is needed because the distance from every mass element to the axis is constant.

13 Uniform disk about its symmetry axis

Consider a thin uniform disk of radius R and mass M.

Its surface density is

(61)

Divide the disk into thin circular rings of radius r and width dr.

The ring area is

(62)

Thus

(63)

The contribution to moment of inertia is

(64)

Therefore

I	= ∫ 0Rr2σ 2πr dr	(65)
	= 2πσ ∫ 0Rr3 dr	(66)
	= 2πσ.	(67)

Substitute

(68)

(69)

The same result holds for a uniform solid cylinder about its symmetry axis because each cross section perpendicular to the axis has the same radial mass distribution.

14 Why the ring has larger inertia than the disk

For equal mass and equal outer radius,

Iring	= MR2,	(70)
Idisk	= MR2.	(71)

Therefore

(72)

The ring places all of its mass at the maximum radius.

The disk places much of its mass closer to the axis, where the r² weighting is smaller.

15 Uniform rectangular plate about a perpendicular central axis

Consider a thin rectangular plate of mass M, side lengths a and b, centered at the origin in the xy plane.

Let the rotation axis be the z axis through the center.

The surface density is

(73)

For an element

(74)

the squared perpendicular distance is

(75)

Thus

Iz

= σ ∫ −a∕2a∕2 ∫ −b∕2b∕2(x2 + y2) dy dx.

(76)

Separate the integrals:

Iz

= σ.

(77)

Using

(78)

we obtain

Iz	= σ	(79)
	= (a2 + b2).	(80)

Therefore

(81)

16 Uniform solid cylinder

For a uniform solid cylinder of radius R, length L, and mass M, rotated about its symmetry axis, use cylindrical coordinates.

A thin cylindrical shell of radius r, thickness dr, and length L has volume

(82)

With uniform volume density

(83)

(84)

Then

I	= ∫ 0Rr2 dm	(85)
	= 2πρL∫ 0Rr3 dr.	(86)

Therefore

(87)

The cylinder length cancels because distance to the symmetry axis depends only on radius.

17 Uniform solid sphere about a diameter

For a uniform solid sphere of radius R and mass M, the moment of inertia about any diameter is

(88)

One direct derivation uses thin spherical shells.

A thin shell of radius r and thickness dr has mass

(89)

where

(90)

A thin spherical shell has moment of inertia

(91)

Therefore

I	= ∫ 0Rr2	(92)
	= ∫ 0Rr4 dr	(93)
	= .	(94)

Substituting the density gives

(95)

The shell result itself can be derived from a surface integral. It is quoted here to keep the solid-sphere calculation focused on the radial assembly of shells.

18 Reference values

For commonly encountered uniform bodies:

These formulas should not be memorized without understanding the axis and mass distribution that define them.

19 Radius of gyration

Any moment of inertia can be written as

(96)

where k_g is the radius of gyration.

Thus

(97)

The radius of gyration is the radius at which the entire mass could be concentrated as a point mass while preserving the same moment of inertia about the chosen axis.

For a uniform disk,

(98)

so

(99)

For a thin ring,

(100)

20 Physical interpretation of radius of gyration

The relation

(101)

shows that k_g is a root-mean-square distance from the rotation axis.

It is not the average radius.

It weights larger radii more strongly because the squared distance is averaged.

21 Scaling of moment of inertia

Suppose two geometrically similar uniform bodies have the same shape but different characteristic size L.

Their moments of inertia have the form

(102)

where C is a dimensionless shape factor.

For example:

Cring	= 1,	(103)
Cdisk	= ,	(104)
Csphere	= .	(105)

If mass is held fixed and all linear dimensions are doubled,

(106)

If density is held fixed instead, mass itself scales with volume:

(107)

Then

(108)

for geometrically similar three-dimensional bodies of fixed density.

22 Composite bodies

If a body consists of several nonoverlapping pieces about the same axis,

(109)

Every component inertia must be evaluated about the same final axis.

A missing region or cavity can be handled by treating the removed material as a negative mass distribution:

(110)

This method is especially useful once M05-05 provides the parallel-axis theorem for components whose own natural center axes do not coincide with the final rotation axis.

23 Example 2: ring plus disk about a common axis

A thin ring and a uniform disk rotate together about a common symmetry axis.

Let

Mr	= 2.0 kg,	Rr	= 0.40 m,	(111)
Md	= 3.0 kg,	Rd	= 0.20 m.	(112)

The ring inertia is

Ir	= MrRr2	(113)
	= 2.0(0.40)2	(114)
	= 0.320 kg m2.	(115)

The disk inertia is

Id	= MdRd2	(116)
	= (3.0)(0.20)2	(117)
	= 0.060 kg m2.	(118)

Therefore

(119)

24 Example 3: rotational kinetic energy

A uniform disk has

M	= 8.0 kg,	(120)
R	= 0.50 m,	(121)

and rotates about its symmetry axis at

(122)

Its moment of inertia is

I	= MR2	(123)
	= (8.0)(0.50)2	(124)
	= 1.0 kg m2.	(125)

Thus

Krot	= Iω2	(126)
	= (1.0)(12)2	(127)
	= 72 J.	(128)

Therefore

(129)

25 Example 4: radius of gyration of a rod

For a uniform thin rod about a perpendicular central axis,

(130)

Thus

kg	=	(131)
	=	(132)
	= .	(133)

Therefore

(134)

26 Moment of inertia is not the area moment of inertia

Engineering uses another quantity often called a second moment of area or area moment of inertia:

(135)

That quantity is important in beam bending and structural mechanics.

The mass moment of inertia used in rigid body dynamics is instead

(136)

The concepts are mathematically similar but physically different and have different units.

27 Moment of inertia and the center of mass

The center of mass uses the first power of position:

(137)

Moment of inertia uses squared perpendicular distance:

(138)

Thus the center of mass tells where the mass is balanced, while moment of inertia tells how strongly the mass is spread away from an axis.

Two bodies can have the same mass and the same center of mass but very different moments of inertia.

28 Common mistakes

1. Quoting a moment-of-inertia formula without specifying the axis.
2. Using distance from the origin instead of perpendicular distance to the rotation axis.
3. Forgetting the square on r_⊥.
4. Treating moment of inertia as depending only on total mass.
5. Using the same I for different parallel or differently oriented axes.
6. Confusing moment of inertia with torque or angular momentum.
7. Using dm = λdx when the density is not uniform without accounting for λ(x).
8. Using surface density for a three-dimensional body or volume density for an ideal thin lamina.
9. Forgetting the Jacobian or geometric factor in dA or dV , such as dA = 2πr dr for a ring element of a disk.
10. Assuming the ring and disk have the same I because they have the same M and R.
11. Confusing radius of gyration with mean radius.
12. Adding component moments of inertia evaluated about different axes.
13. Confusing mass moment of inertia with the area moment of inertia used in structural mechanics.
14. Using the parallel-axis theorem before checking whether the axes are actually parallel. That theorem is developed in M05-05.

29 Practice exercises

1. Four point masses each of mass m lie at the corners of a square of side a. Find the moment of inertia about an axis perpendicular to the square through its center.
2. Three point masses have (m,r_⊥) pairs (1 kg, 1 m), (2 kg, 2 m), and (3 kg, 0.5 m). Find I.
3. Derive the moment of inertia of a uniform thin rod about a perpendicular axis through its center.
4. Derive the moment of inertia of the same rod about a perpendicular axis through one end directly from integration.
5. Derive I = MR² for a thin ring about its symmetry axis.
6. Derive I = MR² for a uniform disk using concentric ring elements.
7. Explain physically why the ring has twice the inertia of a disk with the same M and R.
8. Derive the moment of inertia of a rectangular plate about a perpendicular axis through its center.
9. A solid cylinder has M = 5 kg and R = 0.30 m. Find its moment of inertia about its symmetry axis.
10. A solid sphere has M = 4 kg and R = 0.20 m. Find its moment of inertia about a diameter.
11. Find the radius of gyration of a thin ring, uniform disk, and solid sphere about their standard symmetry axes.
12. A composite rotor consists of a disk and a coaxial ring. Find its total moment of inertia in terms of the component masses and radii.
13. For geometrically similar three-dimensional bodies of fixed density, show that I ∝ L⁵.
14. Explain how two bodies can have the same mass and center of mass but different moments of inertia.
15. A body has I = 6 kg m² and M = 24 kg. Find its radius of gyration.

30 Summary

For discrete masses,

(139)

For a continuous body,

(140)

The appropriate mass elements are

dm = λds	for a line distribution,	(141)
dm = σ dA	for a surface distribution,	(142)
dm = ρdV	for a volume distribution.	(143)

Moment of inertia enters fixed-axis rotational energy and angular momentum:

Krot = Iω2,		(144)
Lz = Izω.		(145)

The radius of gyration is

(146)

For geometrically similar bodies,

(147)

The next article, M05-05, develops the parallel-axis and perpendicular-axis theorems, which allow many new moments of inertia to be obtained from known central-axis results.

References

References