Moment of Inertia and its Theorems

Moment Of Inertia

In translational motion, the mass of an object is a measure of its inertia – its resistance to changes in its linear motion (acceleration). For rotational motion, there is an analogous concept called moment of inertia ($I$). Moment of inertia is a measure of an object's rotational inertia – its resistance to changes in its rotational motion (angular acceleration).

Unlike mass, which is a fixed property of an object, the moment of inertia of a rigid body depends not only on its mass but also on how that mass is distributed relative to the axis of rotation. The same object can have different moments of inertia about different axes.

Consider a system of particles rotating about a fixed axis, or a rigid body rotating about a fixed axis. When a torque is applied, the body undergoes angular acceleration. The relationship between the applied torque ($\vec{\tau}$) and the resulting angular acceleration ($\vec{\alpha}$) for a rigid body rotating about a fixed axis is given by the rotational analogue of Newton's Second Law:

$ \vec{\tau}_{net, ext} = I \vec{\alpha} $

Here, $I$ is the moment of inertia about the axis of rotation. This equation highlights that for a given net external torque, a larger moment of inertia results in a smaller angular acceleration, indicating greater resistance to changes in rotational motion.

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Definition for a System of Discrete Particles

For a system of $n$ discrete particles with masses $m_1, m_2, ..., m_n$ rotating about a fixed axis, with each particle at a perpendicular distance $r_i$ from the axis of rotation, the moment of inertia ($I$) of the system about that axis is defined as the sum of the product of the mass of each particle and the square of its perpendicular distance from the axis:

$ I = m_1 r_1^2 + m_2 r_2^2 + ... + m_n r_n^2 $

This can be written using summation notation:

$ I = \sum_{i=1}^{n} m_i r_i^2 $

where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation. Particles located on the axis of rotation ($r_i=0$) do not contribute to the moment of inertia about that axis.

The units of moment of inertia are kilograms-metre squared (kg·m$^2$). It is a scalar quantity.

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Definition for a Continuous Mass Distribution (Rigid Body)

For a rigid body, which is a continuous distribution of mass, the summation is replaced by integration. If $dm$ is an infinitesimal mass element located at a perpendicular distance $r$ from the axis of rotation, the moment of inertia ($I$) of the body about that axis is given by:

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

The integral is taken over the entire volume or extent of the body. The calculation of moment of inertia for continuous bodies often involves multi-dimensional integration and depends on the shape of the body and the position and orientation of the axis of rotation.

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Radius of Gyration

The moment of inertia of a body about an axis can also be expressed in terms of its total mass $M$ and a quantity called the radius of gyration ($k$).

The radius of gyration is defined as the perpendicular distance from the axis of rotation to a point where, if the entire mass of the body were concentrated, it would have the same moment of inertia as the actual body about the given axis.

$ I = M k^2 $

So, $ k = \sqrt{\frac{I}{M}} $. The units of radius of gyration are metres (m).

The radius of gyration depends on the distribution of mass and the axis of rotation, just like the moment of inertia.

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Moment of Inertia for Common Shapes

Calculating the moment of inertia using integration requires knowledge of calculus and the mass distribution. Here are the moments of inertia for some common rigid body shapes about specific axes (assuming uniform density):

Shape	Axis of Rotation	Moment of Inertia (I)
Thin Ring / Hoop (Mass M, Radius R)	Axis through centre, perpendicular to plane	$MR^2$
Solid Cylinder / Disc (Mass M, Radius R)	Axis through centre, perpendicular to plane	$\frac{1}{2}MR^2$
Solid Sphere (Mass M, Radius R)	Any axis through centre	$\frac{2}{5}MR^2$
Thin Spherical Shell (Mass M, Radius R)	Any axis through centre	$\frac{2}{3}MR^2$
Thin Rod (Mass M, Length L)	Axis through centre, perpendicular to rod	$\frac{1}{12}ML^2$
Thin Rod (Mass M, Length L)	Axis through one end, perpendicular to rod	$\frac{1}{3}ML^2$

Notice that for shapes like the ring and solid disc, even with the same mass and outer radius, the moment of inertia is different about the same axis. This is because the mass distribution is different – in the ring, all mass is concentrated at radius R, while in the disc, the mass is distributed from the centre to R. The disc has less mass farther from the axis, resulting in a smaller moment of inertia.

Theorems Of Perpendicular And Parallel Axes

Calculating the moment of inertia for an extended body about an arbitrary axis can be complex. However, there are two useful theorems that can simplify these calculations if the moment of inertia about a specific axis is already known. These are the Theorem of Perpendicular Axes and the Theorem of Parallel Axes.

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Theorem Of Perpendicular Axes

The Theorem of Perpendicular Axes is applicable only to plane laminar bodies (thin, flat bodies) that lie entirely in a plane.

The theorem states:

The moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes lying in its plane and intersecting each other at the point where the perpendicular axis passes through the plane.

Let the plane laminar body lie in the xy-plane. Let the axis perpendicular to the plane be the z-axis, passing through a point O in the plane. Let $I_z$ be the moment of inertia about the z-axis. Consider two mutually perpendicular axes, the x-axis and the y-axis, lying in the plane of the body and also passing through the same point O.

The theorem states:

$ I_z = I_x + I_y $

where $I_x$ and $I_y$ are the moments of inertia about the x-axis and y-axis respectively.

Proof: Consider a particle of mass $m_i$ located at coordinates $(x_i, y_i)$ in the xy-plane. The perpendicular distance of this particle from the x-axis is $y_i$, and from the y-axis is $x_i$. The perpendicular distance from the z-axis (passing through the origin O) is $r_i = \sqrt{x_i^2 + y_i^2}$.

The moment of inertia about the x-axis is $I_x = \sum_i m_i y_i^2$.

The moment of inertia about the y-axis is $I_y = \sum_i m_i x_i^2$.

The moment of inertia about the z-axis is $I_z = \sum_i m_i r_i^2 = \sum_i m_i (x_i^2 + y_i^2)$.

$ I_z = \sum_i m_i x_i^2 + \sum_i m_i y_i^2 $

$ I_z = I_y + I_x $

This proves the theorem. This theorem is very useful for finding the moment of inertia about an axis perpendicular to a thin plate if the moments of inertia about two perpendicular axes in the plane are known, or vice versa.

Example: For a uniform thin disc of radius R and mass M, the moment of inertia about an axis through its centre and perpendicular to its plane (the z-axis) is $I_z = \frac{1}{2}MR^2$. By symmetry, the moment of inertia about any diameter (an axis in the plane passing through the centre, say the x-axis) is the same as the moment of inertia about any other diameter (say the y-axis). So, $I_x = I_y$. Using the perpendicular axis theorem, $I_z = I_x + I_y = 2I_x$. Therefore, $I_x = I_y = \frac{I_z}{2} = \frac{1}{2} (\frac{1}{2}MR^2) = \frac{1}{4}MR^2$. The moment of inertia of a thin disc about a diameter is $\frac{1}{4}MR^2$.

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Theorem Of Parallel Axes

The Theorem of Parallel Axes is applicable to any rigid body (not just plane laminar bodies) and relates the moment of inertia about any axis to the moment of inertia about a parallel axis passing through the centre of mass.

The theorem states:

The moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis passing through its centre of mass plus the product of the total mass of the body and the square of the perpendicular distance between the two parallel axes.

Let $I$ be the moment of inertia of a body about an arbitrary axis. Let $I_{CM}$ be the moment of inertia of the same body about a parallel axis that passes through the body's centre of mass (CM). Let $d$ be the perpendicular distance between these two parallel axes.

The theorem states:

$ I = I_{CM} + M d^2 $

where $M$ is the total mass of the body.

Proof: Consider the origin of a coordinate system at the centre of mass (CM) of the body. Let the arbitrary axis about which we want to find the moment of inertia be parallel to the z-axis and pass through a point O at position vector $\vec{d}$ from the CM. Let the z-axis through the CM be the axis for $I_{CM}$. The distance $d$ is the perpendicular distance from the origin to the arbitrary axis.

Consider a particle of mass $m_i$ at position vector $\vec{r}_i'$ relative to the CM. The position vector of this particle relative to the arbitrary point O is $\vec{r}_i = \vec{r}_i' + \vec{d}$. The perpendicular distance of $m_i$ from the arbitrary axis is $r_{i,perp}$. The moment of inertia about the arbitrary axis is $I = \sum_i m_i r_{i,perp}^2$.

This requires a more rigorous vector derivation using the perpendicular components. Let the arbitrary axis be along the z-direction passing through $\vec{d} = d_x\hat{i} + d_y\hat{j}$. A particle is at $\vec{r}_i' = x_i'\hat{i} + y_i'\hat{j} + z_i'\hat{k}$ relative to the CM. Its position relative to the point $\vec{d}$ on the arbitrary axis is $\vec{r}_i = \vec{r}_i' - \vec{d} + z_i'\hat{k}$ (considering the point on the axis at the same z-level as the particle for perpendicular distance). No, simpler is to use the general definition $I = \int r^2 dm$. Let the axis through CM be z-axis. $I_{CM} = \int (x'^2 + y'^2) dm$, where $x', y'$ are coordinates relative to CM. Let the parallel axis be shifted by distance $d$ in the x-direction, so its equation is $x=d, y=0$. The perpendicular distance of a particle $(x', y', z')$ from this axis is $r_{perp}^2 = (x' - d)^2 + y'^2 = x'^2 - 2x'd + d^2 + y'^2$.

$ I = \int r_{perp}^2 dm = \int (x'^2 + y'^2 + d^2 - 2x'd) dm $

$ I = \int (x'^2 + y'^2) dm + \int d^2 dm - \int 2x'd dm $

$ I = I_{CM} + d^2 \int dm - 2d \int x' dm $

$ I = I_{CM} + M d^2 - 2d \int x' dm $

The term $\int x' dm$ is related to the x-coordinate of the centre of mass relative to the origin at the CM. By definition, $X_{CM} = \frac{1}{M}\int x' dm$. Since the origin is at the CM, $X_{CM} = 0$. Thus, $\int x' dm = 0$. Similarly $\int y' dm = 0$. The origin of coordinates is at the CM, so $ \int \vec{r}' dm = 0 $.

$ I = I_{CM} + M d^2 - 2d (0) $

$ I = I_{CM} + M d^2 $

This proves the theorem of parallel axes. It is valid for any rigid body and any axis, provided $I_{CM}$ is calculated about a parallel axis passing through the CM.

Example: For a uniform thin rod of length L and mass M, the moment of inertia about an axis through its centre (CM) and perpendicular to the rod is $I_{CM} = \frac{1}{12}ML^2$. To find the moment of inertia about a parallel axis through one end of the rod, the distance between the two axes is $d = L/2$. Using the parallel axis theorem:

$ I_{end} = I_{CM} + M d^2 $

$ I_{end} = \frac{1}{12}ML^2 + M \left(\frac{L}{2}\right)^2 $

$ I_{end} = \frac{1}{12}ML^2 + M \frac{L^2}{4} $

$ I_{end} = \left(\frac{1}{12} + \frac{3}{12}\right)ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2 $

This matches the known result for a rod rotating about its end.