# Mass/Inertia Scalars#

**Mass**: measure of amount of materia in a body. Units of measurement: \(kg\), \(lb\).**Mass center**: consider a set of particles as shown below which together make up the system of particles \(S\):

The \(i^{\text{th}}\) particle has a mass \(m_i\)

\(S^*\) is a fictitious particle such that:

This fictitious particle is called **mass center**. So, how does one locate the mass centre from a point \(O\)?

\({\bf r}^*\); position vector from \(O\) to \(S^*\)

\({\bf q}_n\); position vector from \(O\) to \(P_n\)

So, the position vector to locate the \(n^{th}\) particle is:

Thus, expanding (25)

For a continuum:

where, \(dm\) is elemental mass that can be obtained from density \(\rho\) and elementar volume \(dV\).

## Composite theorem for mass centre#

where,

\(r^*_i\) is the position vector locating the mass centre of \(S_i\), the \(i^{\text{th}}\) system of particles.

\(m_{s_{i}}\) is the mass of \(i^{\text{th}}\) system.

\(r^*\) is the mass centre of the composite system \(S\).

### Example #1#

**Given**:

\(F\) and \(R\) are the bodies of mass density \(\rho\; kgm^{-2}\) and \(\sigma\;kgm^{-1}\) respectively.

\(P\) is a particle of mass \(m\).

**Find**:

Mass centre of the combined system.

### Example #2#

\(F\) is split into two: \(F_1\) and \(F_2\).

\(m_{F_{1}}\) = \(\rho H_a\), is mass of \(F_1\).

\(m_{F_{2}}\) = \(\rho B_a\), is mass of \(F_2\).

Also,

\(m_{R}\) = \(\sigma L\), is mass of \(R\).

Then,

Similarly,

and

## Inertia scalar#

For a particle \(P\) of mass \(m\), we can define a parameter called the inertia scalar. This is defined relative to an arbitrary point \(O\). There are two such inertia scalars:

### 1. Product of inertia#

#### Notation#

\(I^{P/O}_{ab}\) is the product of inertia of \(P\) along two lines through point \(O\) that are parallel to unit vectors \(\hat{n}_a\) and \(\hat{n}_b\).

**8.5** can be extended for both systems of particles and continua.

#### Product of inertia of system particles#

#### Product of inertia of continua#

Warning

In all cases, \(I_{ab} = I_{ba}\) because the formula relies on the dot product of vectors.

### 2. Moment of inertia#

#### Notation#

\(I^{P/O}_{aa}\) is the moment of inertia of P about a line through point \(O\) which is parallel to the unit vector \(\hat{n}_a\).

**8.8** can be extended to both systems of particles and continua.

#### Moment of inertia of system of particles#

#### Moment of inertia of system of particles#

#### Example#

\(P\) is a particle of mass \(m\).

\(\hat{n}_x,\;\hat{n}_y,\;\hat{n}_z\) are unit vectors that are mutually orthogonal.

\(\vec{r} = x\hat{n}_x + y\hat{n}_y + z\hat{n}_z\)

**Find**:

Solution

## From inertia scalars to inertia matrix#

From the previous example, we now have some insight that we will be interested in computing the moments of inertia and products of inertia about a set of unit vectors that make up a reference frame.

For this discussion, we assume that the unit vectors are: \(\hat{n}_x,\;\hat{n}_y,\;\hat{n}_z\).

The inertia scalars can be used to define a square matrix called the inertia matrix.

**Notation**:

\(\left[I\right]^{S/O}\) is the inertia matrix of \(S\), a system of particles about the point \(O\).

The diagonal elements of this matrix are the moments of inertia.

The off-diagonal elements are the products of inertia.

So, the inertia matrix is represented as:

Warning

In all cases, \(I_{ab} = I_{ba}\) because the formula relies on the dot product of vectors.

Rigid body/ continua: The inertia scalars of a rigid body can also be arranged into an inertia matrix.