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Chapter 5: Problem 1

Let \(R=(O, \mathcal{T})\) be a rest frame of a rigid body. Denote the entries in the corresponding inertia matrix by \(\mathcal{J}_{i j}\). Show that the moment of inertia about an axis through \(O\) in the direction of a unit vector with components \(x_{i}\) is \(\mathcal{J}_{i j} x_{i} x_{j} .\)

Short Answer

Expert verified
Answer: The expression is \(I = \mathcal{J}_{11}x_1^2+\mathcal{J}_{22}x_2^2+\mathcal{J}_{33}x_3^2+2\mathcal{J}_{12}x_1x_2+2\mathcal{J}_{13}x_1x_3+2\mathcal{J}_{23}x_2x_3\).

Step by step solution

01

Define the given variables

We are given a rigid body in a rest frame \(R=(O, \mathcal{T})\) with the inertia matrix \(\mathcal{J}_{i j}\). An axis passes through the point O, in the direction of a unit vector with components \(x_i\). The goal is to find the moment of inertia about this axis.
02

Recall the formula for the moment of inertia

The moment of inertia about an axis can be found using the following formula: \(I = \mathcal{J}_{i j}x_i x_j\) Here, \(I\) is the moment of inertia, and the indices \(i\) and \(j\) run from 1 to 3, representing the three dimensions.
03

Apply the formula to the given problem

We can now apply the formula for the moment of inertia to our given problem. Since \(x_i\) are the components of a unit vector in the direction of the axis, we can write: \(I = \mathcal{J}_{i j}x_i x_j\)
04

Expand the expression for \(I\)

To find the moment of inertia, we need to compute the multiplication of the inertia matrix elements with the components of the unit vector squared. The indices \(i\) and \(j\) run from 1 to 3: \(I = \mathcal{J}_{11}x_1^2+\mathcal{J}_{22}x_2^2+\mathcal{J}_{33}x_3^2+2\mathcal{J}_{12}x_1x_2+2\mathcal{J}_{13}x_1x_3+2\mathcal{J}_{23}x_2x_3\)
05

Interpret the result

The moment of inertia about an axis through point O in the direction of a unit vector with components \(x_i\) is given by the expression we derived: \(I = \mathcal{J}_{11}x_1^2+\mathcal{J}_{22}x_2^2+\mathcal{J}_{33}x_3^2+2\mathcal{J}_{12}x_1x_2+2\mathcal{J}_{13}x_1x_3+2\mathcal{J}_{23}x_2x_3\) This expression shows how the moment of inertia depends on the components of the inertia matrix and the unit vector in the direction of the axis.

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Most popular questions from this chapter

The surface of a rigid body is an ellipsoid with equation $$ A x^{2}+B y^{2}+C z^{2}=k^{2} $$ in a rest frame with origin \(O\) at the centre of mass and axes aligned with the principal axes at the centre of mass; \(A, B\), and \(C\) are the principal moments of inertia at \(O\) and \(k\) is a constant. Show that if the body rotates freely about \(O\), with \(O\) fixed relative to an inertial frame, then the two tangent planes to the surface at the intersection points between the surface and the instantaneous axis are fixed relative to the inertial frame. Deduce that the body moves as if it were rolling between these two planes.

Show that the inertia matrix at the centre of any uniform Platonic solid is a multiple of the identity matrix.

Show that the kinetic energy of a uniform rod of mass \(m\) is $$ T=\frac{1}{6} m(\boldsymbol{u} \cdot \boldsymbol{u}+\boldsymbol{u} \cdot \boldsymbol{v}+\boldsymbol{v} \cdot \boldsymbol{v}) $$ where \(\boldsymbol{u}\) and \(\boldsymbol{v}\) are the velocities of the two ends.

Show that the inertia matrix at the centre of a uniform solid cube with mass \(m\) and edges of length \(2 a\) is \(\frac{2}{3} m a^{2} I\). Find the principal axes and principal moments of inertia at a vertex.

\(^\)Show that if a line passing through the centre of mass of a rigid body is a principal axis at one point of the line, then it is a principal axis at every point of the line.
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