In this article, we will learn the moment of inertia of Solid or Hollow Sphere, along with examples, calculation, etc. Let’s explore!

## What is Moment of Inertia of Sphere – Solid & Hollow?

Let’s try to understand the moment of inertia of sphere basics.

### Moment of Inertia Solid & Hollow Sphere Basics

The moment of inertia is also known as the second moment of the area, and it can be calculated for various objects having different shapes. Some basics related to solid and hollow sphere’s moment of inertia are stated below:

- The formula for calculating the moment of inertia of a solid sphere and hollow sphere is derived below in the blog.
- The moment of inertia for any object, including spheres, is an important value calculated using the specific formulas for every object to determine the angular momentum of the entity.
- The moment of inertia is specified to a chosen axis of rotation and depends on the mass distribution around that axis of rotation.
- While finding the moment of inertia of a sphere, whether hollow or solid, and other objects, two theorems are essential, they include the parallel axis theorem and perpendicular axis theorem.

### Moment of Inertia of Sphere Solid & Hollow Definition

The moment of inertia of a sphere is defined as the summation of the products from the whole mass of every attached element of the entity and then multiplying them by the square of the particles with reference to its distance from the center.

## Moment of Inertia of Solid Sphere Formula & Derivation

There is one formula to calculate the moment of inertia of a solid sphere (also known as a spherical shell). How to derive the moment of inertia of a solid sphere, let’s see. To derive, we will split the sphere into infinitesimally thin solid cylinders. After that, we will add the moments of extremely little skinny disks in an exceedingly given axis from left to right. We will look at and perceive the derivation in two different ways.

### Consider an Axis passing about its diameter

Let’s derive the moment of inertia for the solid sphere.

We know, the instant of inertia of a disc can be written as,

- I = (½) MR
^{2}

Infinitesimally moment of inertia and mass can be written as

- dl = ½ r
^{2}dm ——- (1)

Infinitesimally mass of think portion can be written as

- dm = ρ dv ———- (2)

Where,

- ρ = density of Infinitesimally disk
- dm = mass of Infinitesimally disk
- dv = volume of Infinitesimally disk
- r = radius of Infinitesimally disk

Volume of sphere can be written as

- V = 4/3π R
^{3}

Volume of Infinitesimally disk can be written as

- dv = π r
^{2}dx ——– (3)

We have got the value of dm by putting ‘dv’ in Equation (2)

- dm = ρ dv
- dm = ρ π r
^{2}dx

Now, putting the value of dm in Equation (1)

- dl = ½ r
^{2}dm - dI = (½) r
^{2}ρ π r^{2}dx - dI = (½) ρ π r
^{4}dx ———– (4)

Consider a solid sphere with a radius of R and a mass of M. Consider a thin, radiused circular slice y^{2} = R^{2} – x^{2}

y^{2} = R^{2} – x^{2} ——(5)

We have got the value of r and the same needs to be put in Eqn (4);

- dI = (½) ρ π r
^{4}dx - dI = ½ ρπ (R
^{2}– x^{2})^{2}dx

By Integrating,

- ∫dI = ∫
_{-R}^{+R}½ ρπ (R^{2}– x^{2})^{2}dx - I = ½ ρ π ∫
_{-R}^{+R}(R^{2}– x^{2})^{2}dx - I = ½ ρ π x 2 ∫
_{0}^{2R}(R^{2}– x^{2})^{2}dx - I = ρ π ∫
_{0}^{2R}(R^{4}– 2R^{2}x^{2}+ x^{2}) dx - I = ρ π [R
^{4}x – 2R^{2}x^{3}/3 + x^{5}/5]_{0}^{R} - I = ρ π [(R
^{4}R – 2R^{2}R^{3}/3 + R^{5}/5)-0] - I = ρ π R
^{5 }8/15 - I = 8ρπ R
^{5 }/15 ———– (6)

However, the sphere’s mass is M, density is ρ and volume is V. So, we can write,

- ρ = M/V
- ρ = [M / (4/3) πR
^{3}] [Where, V =(4/3) πR^{3}]

Putting the value of density ρ in Eqn. (6), we get;

- I = 8ρπ R
^{5 }/15 - I = 8 [M / (4/3) πR
^{3}] π R^{5 }/15 - I = ⅖ MR
^{2}

So, the moment of inertia of Solid Sphere, I = ⅖ MR^{2}

### Consider an Axis passing about a tangent

Assume that XY is a tangent at A. The parallel axes theorem states that,

- I
_{T }= I_{A}+ Mh^{2} - I
_{T }= I_{A}+ M (0A)^{2} - I
_{T }= 2/5 MR^{2}+ MR^{2} - I
_{T }= 7/5 MR^{2}

You can learn Moment of inertia of Triangle

## Moment of Inertia of Hollow Sphere Formula & Derivation

There is one more formula to calculate the moment of inertia of a hollow sphere (also known as a spherical shell).

How to derive the moment of inertia of a hollow sphere, let’s see.

### Derivation of Hollow Sphere Moment of Inertia Formula

Let us perceive the hollow sphere formula derivation. We already learned that the moment of inertia of a circle can be written as, I = mr^{2}, Where,

- I = Moment of inertia of circle
- m = mass of circle
- r = radius of circle

As per differential equation, we can write the moment of inertia of a circle as;

- dl = r
^{2}dm

Now, we can write ‘dm’ as below,

- dm = (M/A) dA ———– (1)

Where,

- A: Total area of the hollow sphere which is 4πR
^{2} - dA: Smallet area of thin circular portion

Now, dA can be written as,

- The smallest area of hollow sphere = length x width
- The smallest area of hollow sphere = circumference x thickness
- dA = 2πr x Rdθ
- dA = 2πrRdθ ————– (2)

Where,

- 2πr: Circumference of the ring
- R dθ: Thickness

Arc length, S = R θ

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

Now, sinθ = r/R or r = R sinθ

- dA = 2πrRdθ
- dA = 2π. Rsinθ. Rdθ
- dA = 2πR
^{2}sinθ dθ ——— (3)

Now, putting the value of DA in equation in (1),

- dm = (M/A) dA
- dm = (M/A) 2πR
^{2}sinθ dθ - dm = (M/4πR
^{2}) 2πR^{2}sinθ dθ - dm = (M/2) sinθ dθ

Now, as per Equation (1),

- dl = r
^{2}dm - dl = r
^{2}(M/2) sinθ dθ - dl = (Rsinθ)
^{2}(M/2) sinθ dθ - dl = MR
^{2}/2 sin^{3}θ dθ - dl = MR
^{2}/2 sin^{2}θ sinθ dθ - dl = MR
^{2}/2 (1-cos^{2}θ) sinθ dθ - dl = MR
^{2}/2 (1-cos^{2}θ) sinθ dθ —— (4)

Now, Integrating the equation,

- I = MR
^{2}/2 ∫_{0}^{π}(1-cos^{2}θ) sinθ dθ

Let’s consider, cosθ = u, we’ll get;

- I = MR
^{2}/2 ∫_{0}^{π}(1-u^{2}) sinθ dθ - I = MR
^{2}/2 ∫_{1}^{-1}(u^{2}-1) du - I =(MR
^{2}/2) {[ u^{3}/3]_{1}^{-1}– [u]_{1}^{-1} - I =(MR
^{2}/2) {(1/3)[ (-1)^{3}-1^{3}] – [-1-1]} - I = (MR
^{2}/2){[-1-1]/3 – [-2]} - I =(MR
^{2}/2) { [(-2/3) +2]} - I = (MR
^{2}/2) x (4/3) - I = (2/3) MR
^{2}

So, the moment of inertia of Hollow Sphere, I = 2/3 MR^{2}

## Moment of Inertia of Sphere Calculation

### Question#1 Determining the moment of inertia of a hollow sphere

**Problem**: Determine the moment of inertia of a hollow sphere of mass 55 kg. The sphere has a radius of 0.120 m.

**Solution:** The solution requires the following expression for calculation:

I = (2/3)MR^{2}

Substituting the values in the above equation:

I =(2/3) (55.0 kg)(0.120 m)^{2}

I = (55.0 kg)(0.0144 m^{2})

I =(2/3) (0.792 kg.m^{2})

Hence the moment of inertia of the hollow sphere of mass 55 kg and radius 0.120 m is 0.528 kg.m^{2}.

### Question#2 Calculating the moment of inertia of a solid sphere

**Problem**: A solid sphere has a moment of inertia “I” about its diameter and is recast into identical small 8 spheres. Determine the moment of inertia of a smaller sphere about its diameter.

**Solution:** Moment of Inertia of the solid sphere is given by:

I=2MR^{2}/ 5

Since the sphere is recast into 8 smaller pieces hence the mass is given by: M/8

Moreover, the density will remain the same because both bodies have the same material.

ρ=M/V

where “ρ”, “M”, and “V” show density, mass, and volume, respectively.

V=4πR^{3}/3, where “V” shows volume and “R” is the radius.

Suppose that the mass and radius of the bigger sphere are demonstrated by “M” and “R”.

The bigger sphere is recast into 8 smaller spheres; therefore, mass is M/8, and the supposed radius is demonstrated by “r”.

Density is given by: ρ=M/V

Volume of a sphere having radius “R” is given by: V_{R}= 4πR^{3}/3

For the sphere of mass “M” and radius “R”, the density equation will become: ρ= M/ 4πR^{3}/3

Now the volume of a sphere having radius “r” is given by : V_{r}= 4πr^{3}/3

Now the density for sphere having mass M/8 and radius “r” is given by: ρ= (M/8)/4πr^{3}/3

Now density is same, so we can write,

- ρ= M/ 4πR
^{3}/3 = (M/8)/ 4πr^{3}/3 - 4πr
^{3}= 1/8 X 4πR^{3} - r
^{3}= 1/8 X R^{3} - r
^{3}= R^{3}/8 - r= R/2

Moment of inertia of the solid sphere with its diameter is given by: I = 2 M R ^{2}/ 5

The smaller sphere having mass “M/8” and radius “R/2” has a moment of inertia calculated below:

- I
_{small sphere}= 2 . M/8 . (R/2)^{2 }/5^{ } - I
_{small sphere}= 2 MR^{2}/ 5(32)

As we know, I=2MR^{2}/ 5

I_{small sphere}= I/32

### Calculating the moment of inertia of a system consisting of two balls

Calculate the system’s moment of inertia about the rotation axis AB shown in the diagram. The system comprises two balls, X and Y having masses 500 g and 700 g, respectively. r_{x }and r_{y} are 0.2 and 0.3 m, respectively.

Solution: AB is the rotation axis

m_{X} = 700 grams = 0.7 kg

m_{Y} = 500 grams = 0.5 kg

r_{X }= 0.2 m

r_{Y }= 0.3 m

- I = m
_{X}r_{X}^{2}+ m_{Y}r_{Y}^{2} - I = (0.7)× (0.2)
^{2}+ (0.5)× (0.3)^{2} - I = (0.7) x (0.04) + (0.5) x (0.09)
- I = 0.028 + 0.045 I = 0.073 kg m
^{2}

## Conclusion

Hence, we have learned the basics of the moment of inertia of the sphere and how to calculate this value through sample calculation. Our YouTube Refer our YouTube Videos