The Moment of Inertia is otherwise called the Mass Moment of Inertia, or Rotational Inertia, angular Mass of a rigid body, is a quantity, which determines the torque required for a desired angular Acceleration around a Rotational Axis; similar to how the Mass determines Force needed for the desired Acceleration. It completely depends on the Mass distribution of the body and the Axis chosen, with larger Moments requiring more torque to change the rate of Rotation of the body.

It is an extensive or additive property: for a point Mass, simply, the Moment of Inertia is simply the Mass times the square of the perpendicular distance to the Axis of the Rotation. The Moment of Inertia that belongs to a rigid composite system is given by the sum of Moments of Inertia of its component subsystems (all taken about the same Axis).

### Moment of Inertia of a Hollow Sphere

The Moment of Inertia of a Hollow Sphere, otherwise called a spherical shell, is determined often by the formula that is given below.

Formula: $I = \dfrac{2}{3} MR^2$,

Where:

I is the moment of inertia

M is the mass of the hollow sphere

R is the outer radius of the hollow sphere

2. Moment of inertia about an axis passing through the center and perpendicular to a diameter:

Formula: $I = \dfrac{2}{5} MR^2$,

Where:

I is the moment of inertia

M is the mass of the hollow sphere

R is the outer radius of the hollow sphere

Let’s calculate the Moment of Inertia of a Hollow Sphere with a Radius of 0.120 m, a Mass of 55.0 kg

Now, to solve this, we need to use the formula which is;

$I = \dfrac{2}{3} MR^2$

Substituting the values, we get,

$I = \dfrac{2}{3} (55.0 kg)(0.120 m)^2$

$I = \dfrac{2}{3} (55.0 kg)(0.0144 m^2)$

$I = \dfrac{2}{3} (0.792 kg m^2)$

$I = 0.528 kg m^2$

Hence, the value of the Moment of Inertia of the Hollow Sphere is 0.4181 kg.m^{2}.

### Derivation of Hollow Sphere Formula

Let us understand the Hollow Sphere formula derivation.

Before going to derive the formula, let us recall or consider the Moment of Inertia of a circle which is given by,

I = mr^{2}

Applying the differential analysis, we get;

dl = r^{2} dm

Now, we have to find the dm value with the formula,

dm = dA

Where A is the total surface area of the shell, which is given as 4πR2, and dA is the area of the ring that is formed by differentiation and is expressed as;

dA = R dθ × 2πr

Where R dθ is the thickness and 2πr is the circumference of the ring.

It is to make a note that, we get R dθ from the equation of arc length, S = R θ

The next step involves relating r with θ.

Considering the above diagram, we will see the right angle triangle with an angle of θ.

We get,

sin θ = = r = R sinθ

Now, dA becomes as below.

dA = 2πR^{2}sinθ dθ

Substituting the equation for dA into dm, we get,

dm = dθ

Now, we will substitute the above equation and for ‘r’ into the equation for ‘dI.’ Then we get;

dm = sin^{3} θ dθ

Then, integrating within the limits of 0 to π radians from one end to another, we get;

I = sin^{3} θ dθ

Now, we need to split sin^{3}θ into two, because it depicts the case of integral of odd powered trigonometric functions. So, we get;

I = sin^{2} θ sin θ dθ

However, normally, sin^{2} θ is given as sin^{2} θ = 1- cos^{2} θ.

Now,

I = (1- cos^{2} θ) sin θ dθ

We use substitution after this, where u = cos θ, we will get;

I = u^{2} – 1 du

We have to carry out the integration as:

I = u^{2} – 1 du,

Here, the integral of u^{2} du = u and the integral of 1 du = u

Substituting the values, we get.

I = {1-1 – u1-1

I = {[ (-1)3 -13] − − 1−1}

I = {– −2}

I = { +2}

I = {}

I = x

I = MR^{2}

### Examples on How to Solve the Moment of Inertia of a Hollow Sphere

Let us know how to calculate the Moment of Inertia of a Hollow Sphere by using the problem given below.

### Problem

If a Hollow Sphere has an inner Radius of 12 cm, an outer Radius of 18 cm, the Mass of 15 kg, what is the Moment of Inertia of a Sphere (Rotational Inertia of Hollow Sphere) of the Sphere of an Axis passing through its center?

Solution

The volume of a Sphere is given by V =\[\frac{4}{3}\]πR3

Mass = (density)(volume) = ρV

Mass of the Hollow Sphere = (Mass of the solid outer Sphere) - (Mass of the solid inner Sphere)

mHollow = ρVo−ρVi

15kg = ρ[Vo−Vi]

15=ρ[(\[\frac{4}{3}\] π 0 . 183) − (\[\frac{4}{3}\] π 0 . 123)]

(or)

15 = ρ π \[\frac{4}{3}\](0.183 − 0 .123)

ρ= 872.56\[\frac{kg}{m^{3}}\]

The Mass Moment of the Inertia of a solid Sphere about its centroidal Axis is given by:

I= \[\frac{2}{5}\]m R2 = 25(ρV) R2

IHollow = Iouter − Iinner

IHollow = ( \[\frac{2}{5}\] (ρVo) R02) − ( \[\frac{2}{5}\] (ρVi)Ri2)

Where Vo = \[\frac{4}{3}\] π0.183, and Vi = \[\frac{4}{3}\] π 0.123

Problem : Calculate the Moment of Inertia of a Hollow Sphere of Mass M and and diameter D about its diameter.

Solution : The formula for Moment of Inertia of a Hollow Sphere of Mass M and diameter D about its diameter is \[\frac {2MR^2} {3}\].

Now as R=D/2, on substituting the value of R in the above formula,

We get,

I = \[\frac{2M(\frac{D}{2})^2}{3}\]

I = \[\frac {2MD^2} {4x3}\] = \[\frac {MD^2} {6}\]

Therefore, the answer is \[\frac {MD^2} {6}\]

Problem : What is the Moment of Inertia of a Hollow Sphere of Mass M, inner Radius R, outer Radius 2R having uniform Mass distribution about the diameter Axis?

Solution :

The density of Sphere(d)= \[\frac {M} {V}\]

Now, Volume V = 4x/3( (2r)3 - r3)= (4π * 7r3)/3

Suppose that a Hollow Sphere of thickness dx at a distance x from center ( where x lies between r and 2r)

Mass of this Sphere = dm= d(4π2.dx)

dm= \[\frac {3mx^2 dx} {7r^3}\]

Now, the Moment of Inertia of the elemental Hollow Sphere is dI. So,

I = \[\frac {2dmx^2} {3}\]

dI = \[\frac {2m} {7r^3}\] x4.dx

On Integration,

I= \[\frac {62} {35}\] mr2

## FAQs on Moment of Inertia of Hollow Sphere

**1. Explain Inertia with an Example.**

Inertia is the body’s tendency to maintain its equilibrium state. Let us understand the concept of inertia with an example.

When a moving bus suddenly stops, our upper body is jolted forward due to the inertia as the upper body wants to stay in motion as before, according to the inertia.

The linear momentum (mass*velocity) of the body became changed and was directly proportional to the force. That is due to the heavily applied bus brak, and the bus would have come to a stop more quickly; our body would have jolted forward more. Therefore, the given rate of change of momentum is directly proportional to the applied force.

**2. If there are Two Spheres as Hollow and Solid, which has Higher Inertia, and Why?**

It cannot be said until we fix the mass. If both the spheres have the same mass, then, both have the same inertia. Because mass is termed as inertia, but, both have a different moment of inertia. Because, the moment of inertia always depends upon the distribution mass of the axis, on which the moment is being calculated.

Because I = M * R^{2}

In the hollow sphere, the mass is distributed at more distance compared to the solid sphere. So, the hollow sphere of the same mass will possess more inertia than that of the solid sphere.

It is totally about the distribution of mass about the axis on which the moment is being taken.

**3. Is Moment of Inertia an important topic for the Exams?**

Yes, Moment of Inertia is an important topic for the Exams. Students can get numerical problems based on this topic which can help students to score high marks in the Exams. Students can understand this concept of Moment of Inertia from the Vedantu notes. The notes are available on Vedantu that are prepared by expert teachers in a very simple and effective language and can help students to understand the formula for solving the numerical based on it. Several Examples based on this topic are also available that can help students to solve the problems without any difficulty. Thus, students can visit the Vedantu for solving different numerical problems based on the topic and score high marks.

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**5. What is Radius of Gyration?**

Radius of gyration is also used for expressing the Moment of Inertia of a body about an Axis. Radius of Gyration is an imaginary distance that is measured from the center at which the area of cross-section is imagined and focused on a point to get the same Moment of Inertia. It is given by k. If you want to further understand this topic, you can study the notes given on Vedantu. The notes given on Vedantu are reliable and help students to easily understand the complex topics and prepare for the Exams. Students can get detailed information on the topic of Moment of Inertia from Vedantu. The topic is well explained in easy language for the students by expert teachers.