Question

Derive the formulae of Moment of inertia of rod,disc,ring,cylinder,hollow sphere and sphere

Answer 1

in the above case I = moment of inertia of thin uniform rod about an axis passing through one end and perpendicular to length

Answer 2

Answer:

The moment of inertia of a hollow sphere or a spherical shell is often determined by the following formula;

I = MR2

We will look at a simple problem to further understand the usage of the formula.

Let us calculate the moment of inertia of a hollow sphere having a mass of 55.0 kg and a radius of 0.120 m.

Now, to solve such problem we need to use the right formula which is;

I = MR2

We will substitute the values,

I = (55.0 kg)(0.120 m)2

I = (55.0 kg)(0.0144 m2)

I = (0.792 kg.m2)

I = 0.4181 kg.m2

The moment of inertia of the hollow sphere is 0.4181 kg.m2.

Hollow Sphere Formula Derivation

We will now understand the derivation of the moment of inertia formula for a hollow sphere.

Moment of Inertia of a Hollow Sphere

First, let us consider or recall the moment of inertia of a circle which is

I = mr2

If we apply differential analysis we get;

dl = r2 dm

We have to find the dm,

dm = dA

Here, A is the total surface area of the shell = 4πR2

dA is the area of the ring formed by differentiation and is expressed as;

dA = R dθ × 2πr

2πr is the circumference of the ring

R dθ is the thickness

Note: We get R dθ from the equation of arc length which is S = R θ

The next step involves relating r with θ.

If we look at the diagram that is given above, we will see that a right angle triangle with angle θ is present.

We get,

sin θ = = r = R sinθ

Now dA becomes:

dA = 2πR2sinθ dθ

If we substitute the equation for dA into dm, we get:

dm = dθ

We will now substitute the equation given above and for r into the equation for dI. We will get;

dm = sin3 θ dθ

Integrating within the limits of 0 to π radians. From one end to another.

We will get;

I = sin3 θ dθ

Now, we need to split the sin3θ into two, as it depicts the case of integral of odd powered trigonometrical functions. We get;

I = sin2 θ sin θ dθ

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

I = (1- cos2 θ) sin θ dθ

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

I = u2 – 1 du

We have to carry out the integration:

I = u2 – 1 du,

Here integral of u2 du = and integral of 1 du = u

If we substitute the values,

I = {[ ]1-1 – [u]1-1

I = {[ (-1)3 -13] – [-1-1]}

I = {[] – [-2]}

I = { +2}

I = {}

I = x

I = MR2