Moment of Inertia of Hollow Sphere

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.m².

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 = mr²

Applying the differential analysis, we get;

dl = r² 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 = 2πR²sinθ 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 = sin³ θ dθ

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

I = sin³ θ dθ

Now, we need to split sin³θ into two, because it depicts the case of integral of odd powered trigonometric functions. So, we get;

I = sin² θ sin θ dθ

However, normally, sin² θ is given as sin² θ = 1- cos² θ.

Now,

I = (1- cos² θ) sin θ dθ

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

I = u² – 1 du

We have to carry out the integration as:

I = u² – 1 du,

Here, the integral of u² 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 = MR²

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