Moment of Inertia of Hollow Sphere

What is the Moment of Inertia?

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.

I = MR2

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 = MR2

Substituting the values, we get,

I = (55.0 kg) (0.120 m)2

= (55.0 kg) (0.0144 m2)

= (0.792 kg.m2)

= 0.4181 kg.m2

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.

[Image to added soon]

  • 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 that 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 trigonometrical 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 – [u]1-1

I = {[ (-1)3 -13] – [-1-1]}

I = {[] – [-2]}

I = { +2}

I = {}

I = x

I = MR2

Example 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) =  ρ

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

FAQs (Frequently Asked Questions)

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 * R2


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.