The moment of inertia, we also call it the angular mass or the rotational inertia, of a rigid body, is the quantity. Furthermore, it can determine the torque that is needed for the desired acceleration regarding a rotational axis. Moreover, it is similar to how mass can determine the requirement of force for the desired acceleration. Similarly, it depends on the mass distribution of the body and the axis chosen, with larger moments demanding more torque for changing the rotation rate of the body. Thus, the moment of inertia of a rigid composite system is the total sum of the moments of inertia of its component subsystems. Learn moment of inertia formula here.

### Example

#### Simple Pendulum

We can measure the moment of inertia by using a simple pendulum. This is because it is the resistance to the rotation that the gravity causes.

**Moment of Inertia Formula Derivation**

The physical object is made of the small particles. The Mass Moment of Inertia of the physical object is expressible as the sum of Products of the mass and square of its perpendicular distance through the point that is fixed (A point which causes the moment about the axis passing through it).

We denote the Mass Moment of Inertia by I

Let’s take consideration of a physical body that has a mass of m.

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It composes of small particles whose masses are \(m_{1}\), \(m_{2}\), \(m_{3}\), etc. respectively.

The perpendicular distance of each particle from the line as shown figure is:

\(k_{1}\), \(k_{2}\), \(k_{3}\), etc.

From the above statement, the Mass Moment of Inertia for the whole body will be as:

I = \(m_{1}(k_{1})^{2} + m_{2}(k_{2})^{2} + m_{3}(k_{3})^{2}\) +…

From the concept of center of mass and center of gravity, the mass of a body that we assume to be concentrated at a point.

The mass at that point will be m and the perpendicular distance of point from the fixed line is k

∴hence

\(m_{1}(k_{1})^{2} + m_{2}(k_{2})^{2} + m_{3}(k_{3})^{2}\) +… = \(mk^{2}\)

I = \(mk^{2}\)

Where k is the radius of gyration.

**Solved Examples on Moment of Inertia Formula**

#### Question

From a uniform circular disc that has a radius R and mass 9 M, a small disc with radius R/3 is removed. Calculate the moment of inertia of the remaining disc about an axis that is perpendicular to the plane of the disc and passes through the center of the disc.

#### Solution:

The moment of inertia of removed part about the axis passing through the centre of mass and perpendicular to the plane of the disc = \(I_{cm} + md^{2}\)

= \([m\times (R/3)^{2}]/2 + m\times [4R^{2}/9]=mR^{2}/2\)

Therefore, the moment of inertia of the remaining part = moment of inertia of the complete disc – moment of inertia of the removed portion

= \(9mR^{2}/2 – mR^{2}/2 = 8mR^{2}/2\)

Therefore, the moment of inertia of the remaining part \((I_{remaining}) = 4mR^{2}\)

#### Question:

The moment of inertia of a hollow sphere of mass M that has internal and external radii R and 2R about an axis passing through its centre and perpendicular to its plane is:

A) 3/2 \(MR^{2}\)

B) 13/32 \(MR^{2}\)

C) 31/35 \(MR^{2}\)

D) 62/35 \(MR^{2}\)

#### Solution:

\(\rho = \frac{m}{v}\)

\(V = \frac{4}{3}\pi ((2R)^{3}-R^{3})\)

= \(\frac{4\times 7}{3}\pi R^{3}\)

\(\rho = \frac{m}{v}\) = \(\frac{3m}{4\times 7\times \pi R^{3}}\)

Considering the element of the thickness dx with the mass dm (dx is at a distance x from the middle)

Mass = dm = \(\rho 4\pi \times ^{2}dx\)

= \(\frac{3m}{4\times 7\times \pi R^{3}}\times 4\pi x^{2}dx = \frac{3}{7R^{3}}mx^{2}dx\)

The moment of inertia for this elemental hollow sphere will be:

dI = \(\frac{2}{3}dmx^{2}\)

dI = \(\frac{2}{7}\left ( \frac{M}{R_{3}} \right )x^{4}dx\)

After integration:

I = \(\int_{R}^{2R}\frac{2}{35}\left ( \frac{M}{R^{3}} \right )x^{5}\)

= \(\frac{62}{35}MR^{2}\).

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