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 \(m_i\)

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

../_images/25.png

This fictitious particle is called mass center.

How does one locate the mass centre from a point \(O\)?

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  • \(r^*\); position vector from \(O\) to \(S^*\)

  • \(q_n\); position vector from \(O\) to \(P_n\)

So, we have,

\[ P_n = r^* - q_n \]

So, expanding 8.1

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For a continuum:

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

Composite theorem for mass centre

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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

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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,

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Similarly,

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and

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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:

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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 \(\hat{n}_b\).

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

Product of inertia of system particles

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Product of inertia of continua

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Warning

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

2. Moment of inertia

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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 \(\hat{n}_a\).

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

Product of inertia of system of particles

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Product of inertia of system of particles

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Example

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

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

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

Find:

\[ I^{P/O}_{xx} \qquad I^{P/O}_{yy} \qquad I^{P/O}_{zz} \]

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:

../_images/251.png

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.