# Moment of Inertia

**Moment of Inertia:**

Moment of inertia is an important topic and appears in most physics problems involving mass in rotational motion. Generally, MOI is used to calculate angular momentum.

Moment of inertia is defined as a large body resisting angular acceleration which is the product of the square of the distance between the mass of each particle and its axis of rotation. Or, more simply, it can be described as the quantity that determines the torque required for a given angular acceleration on the axis of rotation. Moment of inertia is also known as angular mass or rotational inertia. The SI unit of moment of inertia is kg${m}^{2}$. The moment of inertia is usually determined relative to a selected axis of rotation. It mostly depends on the mass distribution around the rotation axis. The MOI varies depending on the axis selected.

**Formula of Moment of Inertia: **

In general, the moment of inertia is expressed in the form I = m × r²

where,

m = sum of mass product.

r = distance from the axis of rotation.

The scale of the moment of inertia is given by the formula M¹ L² T^{0}.

The role of moment of inertia is the same as the role of mass in linear motion. It measures the body's resistance to change in rotational motion. It is constant for a given rigid frame and a given axis of rotation.

Moment of inertia, I =∑m_{i}r_{i}^{2}. . . . . . . . . (1)

Kinetic energy, K = ½ I ω². . . . . . . . . . (2)

**The Moment of Inertia Depends on:**

The moment of inertia depends on the following factors:

- · Material density
- · Body shape and size
- · Rotational axis (distribution of mass relative to the axis)

We can further classify rotating body systems as follows:

- · Discrete (particle system)
- · Continuous (rigid frame)

Moment of inertia of a system of particles

The moment of inertia of the system of particles is given by:

I = ∑ m_{i} r_{i}^{2 }[from equation (1)]

where r_{i} is the perpendicular distance from the axis with the mass i of the i^{th} particle.

**Moment of Inertia of Rigid bodies:**

The moment of inertia of a continuous mass distribution is found using the integration technique. If the system is divided into an infinitesimally small mass element of mass "dm" and if "x" is the distance between the mass element and the axis of rotation, the moment of inertia is:

I = ∫ r²dm . . . . . . (3)

Below is a step-by-step guide to calculating the moment of inertia:

Moment of inertia of a uniform rod about a perpendicular bisector

Moment of Inertia of a Uniform Rod :

Calculation of the Moment of Inertia:

Consider a uniform rod of mass M and length L and the moment of inertia should be calculated for the bisector AB. Origin is 0.

The observed mass element 'dm' is between the origin x and x dx. Since the rod is flat, the mass per unit length (linear mass density) remains constant.

Therefore, the moment of inertia of a uniform rod about a perpendicular bisector (I) = ML^2/(12).

** Moment of Inertia of a circular Ring about its Axis:**

Consider a line perpendicular to the plane of the tire through its center. The radius of the tire is R and the mass is M. All elements are at the same distance from the axis of rotation R.

The linear mass density is constant.

Limits: θ = 0 to 2π includes the entire mass of the tire

∴ I = R² [M/2π] × [θ]^{2π}_{0}

Therefore, the moment of inertia of a circular ring about its axis is (I) = MR².

Note that for one-dimensional bodies, if they are flat, their linear mass density (M/L) remains constant. Similarly, for 2D and 3D, M/A (Area Density) and M/V (Volume Density) remain constant respectively.

Moment of Inertia of a rectangular plate about a line parallel to the edge and passing through the Center:

**Moment of inertia of a rectangular plate**

A mass element can be taken on the axis AB between x and x dx.

Since the plate is flat, M/A is constant.

Limits: The left end of the rectangular plate is x = -l/2 and the whole plate is covered by taking x to x = -l/2 to x = l/2.

Therefore, the moment of inertia of a rectangular plate about a line parallel to the edge and passing through the center is (I) =$\frac{M{I}^{2}}{12}$

Note: If the mass element is chosen parallel to the length of the plate, the moment of inertia would be I =$\frac{M{b}^{2}}{12}$ Moment of inertia of a uniform circular plate about its axis.

**Moment of inertia of a Uniform Circular Plate about its Axis:**

Let the mass of the plate be M and the radius R. The center is at point O and the axis is perpendicular to the plane of the plate. The considered mass element is a thin ring between x and x dx of thickness dx and mass dm.

Since the plane is flat, the surface mass density is constant.

Limits: If we take the range x = 0 to x = R for all mass elements, we cover the entire slab.

$I=\left(\frac{2M}{{R}^{2}}\right){\left[{\frac{x}{4}}^{4}\right]}_{0}^{R}=\left(\frac{2M}{{R}^{2}}\right)\times \frac{{R}^{4}}{4}=\frac{M{R}^{2}}{2}$

Therefore, the moment of inertia of a uniform circular plate about its axis is (I) = $\frac{M{R}^{2}}{2}$

**Moment of Inertia of a Thin Spherical Shell or Uniform Hollow Sphere: **

** Moment of Inertia of a thin spherical shell **

Let M and R be the mass and radius of the sphere, O its center, and OY the given axis. The mass is spread over the surface of the ball and the inside is hollow.

Consider the radii of the sphere at angle θ and angle θ dθ with the axis OY and take an element (thin ring) of mass dm with radius R sin θ when these rays are rotated around OY. The width of this ring is Rdθ and its circumference is 2πRsinθ. Since the hollow sphere is uniform, the surface mass density (M/A) is constant.

Limits: As θ increases from 0 to π, element rings cover the entire spherical surface.

Now, integrating the above equation using the method of substitution, we get,

Therefore, the moment of inertia of a thin spherical shell and a uniform hollow sphere is (I) =$\frac{2M{R}^{2}}{3}$

**Moment of Inertia of a Uniform Solid Sphere**

Let us consider a sphere of radius R and mass M. A thin spherical shell of radius x, mass dm, and thickness dx is taken as a mass element. Volume density (M/V) remains constant as the solid sphere is uniform.

Moment of Inertia for Different Objects:

As we observed from the table above, the moment of inertia depends on the axis of rotation. So far we have calculated the moment of inertia of these objects when the axis passes through their (I cm). Once you have selected two different axes, you will notice that the object tolerates rotation differently. Therefore, the following theorems are useful for finding the moment of inertia about any given axis:

**Parallel axis Theorem and Perpendicular axis Theorem**

**Parallel Axis Theorem:**

The parallel axis theorem says that the moment of inertia of a body about an axis parallel to the body passing through its center is equal to the sum of the moment of inertia between the axis passing through the center of the body and the moment of inertia multiplied by the square of the mass. of the body of the distance between the two axes.

The Formula for the Parallel Axis Theorem:

The parallel axis theorem can be expressed as:

where,

I is the moment of inertia of the body

I_{c} is the moment of inertia about the center

M is the mass of the body

h² is the square of the distance between the two axes

Derivation of the Parallel Axis Theorem:

Let Ic be the moment of inertia of the axis passing through the (AB in the picture) and let I be the moment of inertia around the axis A'B' at a distance h.

Consider a particle of mass m at a distance r from the center of gravity of the body. Then

Distance A'B' = r h

So the above is the formula for the parallel axis theorem. Parallel axis theorem

** Parallel Axis Theorem:**

The rod parallel axis theorem can be determined by finding the moment of inertia of the rod. The moment of inertia of the rod is given by:

The distance between the end of the rod and its center is given as:

Therefore, the bar parallel axis theorem is:

** Perpendicular Axis Theorem: **

The Perpendicular axis theorem says that "The moment of inertia of any plane body around any axis perpendicular to that plane is equal to the sum of the moments of inertia around any two perpendicular axes in the plane of the body intersecting the first axis of the plane."

The formula for the perpendicular axis theorem:

The term transverse axis is used when the body is symmetrical about two of the three axes. If the moment of inertia of two axes is known, the moment of inertia about the third axis can be found using the expression:

In an engineering application, let's say we need to find the moment of inertia of a body, but the body is irregular in shape, in such cases, we can use the parallel axis theorem to get the moment of inertia at a point if we know the center of body weight This is a very useful theorem in space physics, where calculating the moment of inertia of spacecraft and satellites makes it possible to reach the outer planets and even deep space. The vertical axis theorem is helpful in applications where we do not have access to one axis of the body and it is crucial for us to calculate the moment of inertia of the body around that axis.

** Q: **If the moment of inertia of the object on the transverse axis passing through its center of gravity is 50 kg·m²and the mass of the object is 30 kg. What is the moment of inertia of the same object on another axis 50 cm away from and parallel to the axis of flow?

Solution: From the parallel axis theorem,

**Kinematics of rotational motion about a fixed axis **

As we know, rotational motion and translational motion are analogous to each other in every respect. Likewise, the terms we use for rotational motion, such as angular velocity and angular acceleration, correspond to the terms velocity and acceleration in translational motion. In this regard, we see that the rotation of a body around a fixed axis is analogous to the linear motion of a body in translational motion. In this section, we consider the kinematics of a body undergoing rotational motion about a fixed axis.

**Angular Velocity:**

Consider an object rotating about a fixed axis as shown in the figure. Consider a particle P on a rotating object. As the object rotates around an axis passing through O, the particle P moves from one point to another such that the angular displacement of the particle is "Ɵ".

It can be said that at the moment t = 0 the angular displacement of the particle P is 0 and at the moment t its angular displacement is Ɵ. Now the rate of change of angular displacement with time is called the angular velocity of the particle. Mathematically, angular velocity

ω = angular velocity

dì = Change in angular position

dt = time interval

SI unit = Rad/sec

Angular acceleration

The angular acceleration of a particle P is defined as the rate of change of the object's angular velocity. Mathematically, the angular acceleration,

dω = change in angular velocity

dt = time interval

SI unit = Rad/s²

** Kinematic equation of rotational motion **

We see that the kinematic quantities of rotation of object P are angular displacement (Ɵ), angular velocity (ω), and angular acceleration (α). These quantities correspond to the displacement (x), speed (v), and acceleration (a) of the linear motion.

As we know, the kinematic equations of linear motion are given as:

where,

${v}_{0}$ is the initial velocity of the body

a is the acceleration of the body

t is time

s is the movement of the body in the given time t

v is the velocity of the object at time t. Analogous equations of reciprocal motion can be presented as follows:

Where, ${\theta}_{0}$ is the initial angular displacement, ωo is the initial angular velocity, α is the angular acceleration, ω is the angular velocity at any instant t.

where,

${\theta}_{0}$ is the initial angular displacement around the z-axis in radians

Ɵ is the final angular displacement around the z-axis in radians

${\omega}_{0z}$ ωoz is the initial angular velocity around the z-axis in time s-1

${\omega}_{z}$ is the final angular velocity around the z-axis in time s-1

t is time s.

**Angular moment - rotation about a fixed axis **

If we consider an extended object as a system of small particles that make up its body, the rate of change of angular momentum of the system of particles at a given point is equal to the total external torque acting on the system.

The Angular Momentum of an Object Rotating About a Fixed Axis:

Consider an object rotating about a fixed axis as shown in the figure. Consider a particle P in a body rotating about an axis as shown above. The total angular momentum of this system can be expressed as:

Where Pi is the momentum of the particle (which is equal to mv) and ri is the distance of the particle from the axis of rotation.

The total angular momentum contribution of a single particle can be given as:

Here the vector r is equal to OP and we can write OP = OC +CP.

So that we can write,

Here,

where r_{p} is the perpendicular distance of point P from the axis of rotation.

As we can see, the tangential velocity v at P is perpendicular to the vector rp. Using a rule of thumb, we can say that the input direction CP×v is parallel to the axis of rotation. Similarly, we see that the product of vectors OC×V is perpendicular to the axis of rotation.

So that we can write,

The observed object is usually symmetrical about the axis of rotation; therefore, every particle with velocity vi has a corresponding particle with velocity -vi at a given perpendicular distance rp, located in the circle in the diametrically opposite direction. Therefore, the total angular momentum produced by these particles is zero because they cancel each other out.

For such symmetrical objects, the total momentum of the object is given by the formula,

Where I is the moment of inertia of the body, ω is the angular velocity of the body, and gives the direction of the total momentum.

**Conservation of Angular Momentum: **

It is the rotational analog of linear momentum, denoted by l, and the angular momentum of a particle in rotational motion is defined as:

It is the cross product of r, ie. the radius of the circle formed by the rotational movement of the body and the linear momentum of the body p, the magnitude of the cross product of two vectors is always the product of their magnitudes by the sine. of the angle between them, so, in the case of angular momentum, the magnitude is given by the formula,

The ratio of torque to angular momentum

The relationship between torque and angular momentum can be found as follows:

Now notice the first term,

Velocity and linear momentum are in the same direction, so their cross-product is zero.

So the rate of change of angular momentum is Torque.

** Conservation of Angular Momentum - Calculation **

The angular momentum of the system is conserved as long as there is no external net torque on the system, due to the law of conservation of momentum, the Earth has been rotating on its axis since the formation of the Solar System,

There are two ways to calculate the angular momentum of any object, if it is a rotating point object, then our angular momentum is equal to the product of the radius of the object.

If we have an extended object like our Earth, the angular momentum gives the moment of inertia, i.e. how much mass does the object have in motion and how far is it from the center, multiplied by the angular velocity,

But in both cases, as long as there is no net force acting on it, the angular momentum before is equal to the angular momentum after some time, imagine spinning a ball attached to a long string, the angular momentum would give.

Now if we somehow reduce the radius of the ball by shortening the wire as it spins, then r will decrease, now according to the law of conservation of momentum, L should stay the same, there is no way for the mass to change, so the vector v should increase so that the momentum remains constant, that is proof of the conservation of angular momentum.

**Rolling Motion Formula and Diagram **

In our daily life, we observe various moving cars, bicycles, rickshaws etc. All these circular wheels are moving. A body, such as a ball or wheel, or a circular object rolling on a predominantly horizontal surface, experiences rolling motion and has a single point of contact at each instant. In the next discussion, we will discuss rolling motion in detail. Suppose a disc-shaped object rolls across a surface without slipping. In other words, at any moment of time, the part of the plate in contact with the surface is at rest relative to the surface.

Rotational motion is a combination of translational motion and rotational motion. For a song, the movement of the is the translational movement of the body. As the piece rolls, the contact surfaces twist slightly for a moment. Due to this deformation, the limited areas of both bodies touch each other. The general effect of this phenomenon is that the component of the contact force parallel to the surface opposes the motion, causing friction.

**Rotational motion**

Let ${v}_{cm}$be the speed of the disk-shaped body. Since the roller disc would be in the geometric center C, the speed of the body, or speed C, is ${v}_{cm}$, which is parallel to the roller surface. The rotational movement of a body is about its axis of symmetry, so the speed at any point ${P}_{0}$, ${P}_{1}$, or ${P}_{2}$ on the body consists of two parts, the translational speed ${v}_{cm}$ and the rotational movement due to its linear speed vr, where vr. = rω. ω is the angular velocity of the rolling plate. ${v}_{r}$ is perpendicular to the radius vector at some point on the disc relative to the geometric center C. Consider the point P0 on the disc. ${v}_{r}$is directed against ${v}_{cm}$ and at this point ${v}_{r}$ =${R}_{\omega}$, where R is the radius of the plate. Therefore, the condition for the disc to roll without slipping is given by the formula ${v}_{cm}$= ${R}_{\omega}$.

The kinetic energy of such a rolling body is obtained as the sum of the translational and rotational kinetic energies. where,

m is the mass of the body

V_{cm} is the translation speed

I moment of slowness

ω is the angular velocity of the rolling body