# Moment of Inertia

The **moment of inertia** is a physical quantity which tells how easily a body can be rotated about a given axis. It is a rotational analogue of mass. It plays the same role in rotational motion as ‘inertia’ does in translational motion. *Inertia* is the property of matter which resists change in its state of motion. Inertia is a measure of the force that keeps a stationary object stationary, or a moving object moving at its current speed. The larger the inertia, the greater the force that is required to bring some change in its velocity in a given amount of time. Suppose a heavy truck and a light car are both at rest, then intuitively we know that more force will be required to push the truck to a certain speed in a given amount of time than will be needed to push the car to that same speed in the same amount of time..

Similarly, moment of inertia is that property where matter resists change in its state of rotatory motion. The larger the moment of inertia, the greater the amount of torque that will be required to bring the same change in its angular velocity in a given amount of time.

Unlike inertia, moment of inertia depends not only on the mass but also the distribution of mass around the axis about which the moment of inertia is to be calculated. An object can have different moments of inertia about different axes. That is, to rotate an object about different axes with an equal angular acceleration, different torque (or effort) is required.

#### Contents

- Moment of inertia, how intuitive is it?
- General properties and ideas
- Point mass
- Moment of inertia of mass distributions
- Relation of moment of inertia with kinetic energy of the system
- Perpendicular axis theorem
- Center of mass - acceleration
- Center of mass of a collection of point
- Center of a mass distribution
- Parallel axis theorem

## Moment of inertia, how intuitive is it?

Moment of inertia shows the tendency of an object to stay in its state of rotatory motion. If an object has more mass, then it is more difficult to rotate it. Consider two spheres of the same radius: one made up of wood, and the other of iron. Both are at the same distance from the axis of rotation.

Which one is easier to rotate? The one which is made up of wood will be easier to rotate as it is of lesser mass. It requires more efforts for larger mass to set up into the rotation. Thus, moment of inertia depends upon mass.

Consider the two identical objects of the same mass at different distances from the axis of rotation. It is not equally easy to rotate both of them about the same axis of rotation. More efforts are required for the object at a greater distance to accelerate to the same angular velocity. Thus, it can be calculated that moment of inertia is dependent on the distance from the axis. If the mass is farther away from the axis, its moment of inertia is greater.

## Consider a cricket bat, as shown in the diagram below. There are two axes about which the bat can be rotated. About which axis is it easier to rotate? Do we require the same or different amount of torque to produce the same angular acceleration about the two axes? If different, then about which axis is less torque required?

Well, the answer is that it is easier to rotate about axis-2. As the mass moves away from the axis, it becomes more difficult to rotate. Thus, the bat has a different moment of inertia about the axes. The moment of inertia of the bat is less about axis-2 compared to that about axis-1. Therefore, we can say that as the mass moves away from the axis, its moment of inertia increases and it becomes more difficult to rotate.

A metal ring is melted and a solid sphere is made out of it. What happens to the moment of inertia about the vertical axis through the center?

## General properties and ideas

- Moment of inertia is a tensor quantity. It has different values for different axes.
- It depends upon the mass as well as the mass's distribution around its axis.
- A body can have different moments of inertia about different axes.
- It is an inherent property of matter by which it tries to maintain its state of angular motion unless and until it is compelled by external torques.
- It is an extensive (additive) property: the moment of inertia of a composite system is the sum of the moments of inertia of its components' subsystems (all taken about the same axis).

## Point mass

The moment of inertia of a

point mass\(m\) about an axis at a perpendicular distance of \(r\) from it is given by \( mr^2 \).

Therefore if the distance of a point mass from the axis is doubled, then the moment of inertia will be quadrupled. If the mass is doubled, then the moment of inertia will also be doubled.

## Moment of inertia of mass distributions

The moment of inertia of an \(n\)-point mass system \( \{ m_i \}_{i=1}^n \) at perpendicular distances \( \{ r_i \}_{i=1}^n \) from the axis of rotation is given by \(\displaystyle \sum_{i=1}^n m_i r_i^2 \).

The moment of inertia of the masses adds up just as a scalar quantity would.

**Avoid this pitfall:**

It should also be noted that the moment of inertia of a system of particles about an axis is **not** the same as the moment of inertia of the center of mass of the system of particles about the same axis.

## Find the moment of inertia of a point mass system consisting of six equal masses, each of mass \(m\) placed at the corners of a regular hexagon of side length \(a\) about an axis passing through the center of the hexagon and perpendicular to its plane. Find the moment of inertia of mass of \(6m\) placed at the center of mass of the above system.

According to the formula, the moment of inertia is \(\displaystyle I = \sum_{i=1}^6 m_i r_i^2 \). Here all the masses are the same, so \( m_i = m \) for \( i = 1,2,3,4,5,6 \). Also in a regular hexagon, the distances of the all the corners from the center are the same and equal to the side length of the hexagon. Thus, \( r_i = a \) for \( i = 1,2,3,4,5,6 \). Therefore, \[ I = \sum_{i=1}^6 m a^2 = 6ma^2 .\]

**Moment of inertia of continuous mass distribution:**

A continuous mass system can be thought of as a collection of infinite mass particles. A bigger object can be broken down into infinitely small elemental point masses. The total moment of inertia will be the sum of all these particles. An integration helps to add the moments of inertia of all these particles.

Thus, the moment of inertia of a distributed mass system can be written as

\[I = \int dm\, r^2,\]
where \(I\) is the moment of inertia, \(dm\) is the mass of a small element considered on the object, and \(r\) is the distance of the elemental mass from the axis.

Calculate the moment of inertia of a uniform straight rod of mass \(M\)and length \(L\) about an axis passing through one of its ends.

Consider an elemental particle present at a distance \(x\), having a length of \(dx\) from the axis. Now we know that the total mass \(M\) is present in a length of \(L\) and hence the mass of a particle of width \(dx\) would be \(M\cdot dx/L=dm\). We then can simply integrate this as follows:

\[\begin{align} I&=\int_0^Ldm \cdot x^2\\ &=\int_0^L\frac{Mdx}L\times x^2\\ &=\frac ML\times \left[\frac{x^3}{3}\right]_0^L\\ &=\frac{ML^2}{3}. \end{align}\]

As we now know, a body can have different moments of inertia about different axes. Then, is there any relation among these moments of inertia? Yes, the moments of inertia about a few axes can be found using two theorems:

- perpendicular axis theorem
- parallel axis theorem.

## Relation of moment of inertia with kinetic energy of the system

For the particle of mass \(m_i\) at a distance \(r_i\) from the axis, the linear velocity is \(v_i=r_i\omega\). Thus the kinetic energy \((k_i)\) of the motion of the particle is

\[k_i=\dfrac{1}{2}m_iv_i^2=\dfrac{1}{2} m_ir_i^2\omega^2.\]

The total kinetic energy \(K\) of the body is given by the sum of kinetic energies of individual particles (suppose there are \(n\) particles). Hence, we have

\[K=\sum_{i=1}^{n} k_i=\dfrac{1}{2} \sum_{i=1}^{n} \left(m_ir_i^2\omega^2\right).\]

Since \(\omega\) is the angular velocity constant for all particles, we can take it out of the summation to obtain

\[K=\dfrac{1}{2}\omega^2 \sum_{i=1}^{n} \left(m_ir_i^2\right).\]

Since we have defined moment of inertia as \(I=\displaystyle\sum_{i=1}^n m_ir_i^2,\) we have the relation

\[K=\dfrac{1}{2}I\omega^2.\]

## Perpendicular axis theorem

For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object:

\[{I_x} + {I_y} = {I_z}.\]

**Avoid this pitfall:**

Perpendicular axis theorem is valid only for planner objects like a thin disc, rings, triangular plates, etc. For objects like spheres, cylinders, cones, etc., this theorem can not be applied.

Consider a lamina lying in the \(xy\)-plane. The \(z\)-axis passes through the intersection of the \(x\)- and \(y\)-axes and is perpendicular to the plane of lamina:

If we take any point on the lamina of mass \(dm\) at coordinate \((x,y,z),\) then the moment of inertia about the \(x\)-axis can be calculated as

\[{I_x} = \int_{}^{} {dm\,{y^2}}. \qquad (1)\]

Similarly,

\[{I_y} = \int_{}^{} {dm\,{x^2}}. \qquad (2)\]

For the moment of inertia about the \(z\)-axis, let the distance from the \(z\)-axis be \(r = \sqrt {{x^2} + {y^2}},\) then \[{I_z} = \int_{}^{} {dm\,{r^2}}= \int_{}^{} {dm\,({x^2}+{y^2})}. \qquad (3)\]

Adding \((1)\) and \((2)\) and using \((3),\) we get

\[\begin{align} {I_x} + {I_y} &= \int_{}^{} {dm\,{x^2}} + \int_{}^{} {dm\,{y^2}} \\ &= \int_{}^{} {dm\,({x^2} + {y^2})} \\ &= {I_z}. \end{align}\]

To learn about the second theorem, i.e. parallel axis theorem, we first need to learn about the concept of center of mass.

## Center of mass - acceleration

The center of mass of a system is a very special point. It is that point whose acceleration depends only on the external forces. For a system of particles, the acceleration of the center of mass equals net external force divided by the total mass of the system: \[\overrightarrow {{a_\text{cm}}} = \frac{{{{\overrightarrow F }_\text{net,external}}}}{{{m_\text{total}}}}.\]

## Consider the diagram shown below. A block of mass \({m_1}\) is acted upon by a force \(F\) in horizontal direction. If the ground is smooth, then what is the acceleration of the center of mass?

We have \[{a_\text{cm}} = \frac{{{F_\text{net,external}}}}{{{m_\text{total}}}}.\]

Here,

\[{F_\text{net,external}} = F.\]

Therefore,

\[{a_\text{cm}} = \frac{F}{{{m_1} + {m_2}}}.\]

## Center of mass of a collection of point

It's also possible to find the actual center of mass, the center of mass of a collection of points.

Suppose that we have \(N\) particles of mass \({m_1}, {m_2}, \ldots, {m_N}\) at respective coordinates

\[({x_1},{y_1},{z_1}),\,({x_2},{y_2},{z_2}),\ldots,({x_N},{y_N},{z_N}).\]

Then, the \(x\)-coordinate of the center of mass is \(\displaystyle {x_\text{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} +\cdots+ {m_N}{x_N}}}{{{m_1} + {m_2} + \cdots+ {m_N}}}.\)

Similarly, the \(y\)-coordinate of the center of mass is \(\displaystyle {y_\text{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + \cdots+ {m_N}{y_N}}}{{{m_1} + {m_2} +\cdots+ {m_N}}}.\)

Likewise, the \(z\)-coordinate of the center of mass is \(\displaystyle {z_\text{cm}} = \frac{{{m_1}{z_1} + {m_2}{z_2} + \cdots+ {m_N}{z_N}}}{{{m_1} + {m_2} + \cdots + {m_N}}}.\)

## Find the distance of the center of mass of a two-particle system of masses \({m_1}\) and \({m_2}\) from mass \({m_1}\), given that the distance between the two masses is \(r.\)

In the diagram below, let mass \({m_1}\) be at the origin, and mass \({m_2}\) at the coordinate \((r,0).\)

Then, \[{x_{cm}} = \frac{{{m_1}0 + {m_2}r}}{{{m_1} + {m_2}}}\]

and \[{y_{cm}} = \frac{{{m_1}0 + {m_2}0}}{{{m_1} + {m_2}}} = 0.\]Thus the distance of the center of mass from \({m_1}\) is \(\displaystyle \frac{{{m_2}r}}{{{m_1} + {m_2}}}.\)

Note:The distance of the center of mass from \({m_2}\) is \(\displaystyle \frac{{{m_1}r}}{{{m_1} + {m_2}}}.\)

## Center of a mass distribution

A continuous mass distribution contains infinite point mass particles. A center of mass can be defined for such a system of particles with the help of integration. First, break the system into infinite small point masses and then integrate to get the location of the center of mass.

The location of the center of mass of a continuous mass distribution can be calculated as

\[X_\text{cm} = \frac{\int x\, dm}{\int \, dm},\]
where \(dm\) is the mass of an elementary particle, \(x\) is the \(x\)-coordinate of the elementary particle, and \({X_\text{cm}}\) is the \(x\)-coordinate of the center of mass.

Similarly,

\[Y_\text{cm} = \frac{\int y\, dm\,} {\int \, dm}, \quad Z_\text{cm} = \frac{\int z\, dm\,} {\int \,dm} .\]

## Parallel axis theorem

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

If an object has the moment of inertia \({I_\text{cm}}\) about an axis passing through its center of mass, then the moment of inertia \(I\) about another parallel axis at a distance of \(d\) will be

\[I = {I_\text{cm}} + m d^2 .\]

**Cite as:**Moment of Inertia.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/calculating-moment-of-inertia-of-point-masses/