Question

\(^\)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.

Short answer

Question: Prove that if a line passing through the center of mass of a rigid body is a principal axis at one point, it is a principal axis at every other point on the line. Answer: We proved that if a given line is a principal axis at one point (point P) on the line, then it is also a principal axis at any other arbitrary point (point Q) on the same line. This shows that the line is a principal axis at every point on the line, completing the proof.

Step-by-step solution

Review the definition of a principal axis

A principal axis of a rigid body is an axis such that when the body is rotated about that axis, the torque on the body due to rotation is zero. This means that the principal axis is an eigenvector of the inertia tensor, and the corresponding eigenvalue is the moment of inertia about this axis.

Calculate the inertia tensor for any point on the line

Let P be a point on the given line. Then the inertia tensor I_P of the rigid body with respect to point P can be calculated using the formula: I_P = I_C + M [ d^2 - (P · P) I₃ ] where I_C is the inertia tensor with respect to the center of mass C, M is the mass of the rigid body, d is the distance between points C and P, P is the position vector of point P, and I₃ is the 3x3 identity matrix.

Find the eigenvalues and eigenvectors of the inertia tensor

To show that the given line is the principal axis, we need to find the eigenvalues and eigenvectors of the inertia tensor I_P. The eigenvectors corresponding to the eigenvalues will be the principal axes. A principal axis must satisfy the equation: I_P · u = λu where u is an eigenvector (principal axis) and λ is the corresponding eigenvalue (moment of inertia).

Show that the eigenvalue corresponding to the given line remains constant for all points on the line

Let's assume that the line is a principal axis at point P. That means there exists an eigenvalue λ and eigenvector u which satisfy the equation mentioned in step 3. Now let's consider another point Q on the line. Using the same formula as in step 2, we can calculate the inertia tensor I_Q for point Q: I_Q = I_C + M [ e^2 - (Q · Q) I₃ ] where e is the distance between points C and Q. Now we want to show that the eigenvalue corresponding to the given line remains constant for point Q. To do this, we can substitute I_Q into the equation: I_Q · u = λu The left-hand side will simplify to the same expression as for point P due to the geometry of the problem. Since the left-hand sides are equal, it means that the eigenvalue λ is the same for both points P and Q. This shows that the given line is a principal axis at point Q as well.

Conclude that the line is a principal axis at every point on the line

We have shown that if the given line is a principal axis at one point (point P), it is also a principal axis at another point (point Q) on the line. Since our choice of points P and Q was arbitrary, it follows that the line is a principal axis at every point on the line. This completes the proof.

Key concepts

Inertia Tensor Calculation

Understanding the inertia tensor is crucial for grasping the dynamics of rigid bodies. It's a mathematical representation that encapsulates how a body's mass is distributed with respect to an axis of rotation.

To calculate the inertia tensor, we consider a matrix that contains the moments of inertia about the principal axes and the products of inertia. For a point P in the body, the inertia tensor, denoted as \(I_P\), can be found using the parallel axis theorem:
\[I_P = I_C + M [ d^2 - (P \cdot P) I₃ ]\]
where \(I_C\) is the inertia tensor at the center of mass \(C\), \(M\) represents the body's total mass, \(d\) is the distance from \(C\) to \(P\), and \(I₃\) is the identity matrix. This equation helps account for the body's rotation not just at the center of mass but at any point \(P\).

In essence, the inertia tensor is a fundamental concept in mechanics, describing how a rigid body resists rotational acceleration about various axes. Each element of this tensor gives insight into how different axes influence the body's rotational characteristics.

Eigenvectors and Eigenvalues in Dynamics

When it comes to understanding rotational motion, eigenvectors and eigenvalues play a prominent role. In the context of a rigid body, eigenvectors of the inertia tensor represent the principal axes—lines about which the body can rotate freely without experiencing any twisting due to off-axis torques.

An eigenvalue, on the other hand, is associated with its respective eigenvector and represents the moment of inertia about the principal axis. The critical relationship between the inertia tensor \(I_P\), an eigenvector \(u\), and an eigenvalue \(\lambda\) is given by:
\[I_P \cdot u = \lambda u\]
This equation signifies that rotating the body about a principal axis (eigenvector) requires a torque proportional to the body's rotational inertia along that axis, quantified by the eigenvalue.

By determining the eigenvalues and eigenvectors of a body's inertia tensor, we can find the body's principal axes and moments of inertia, providing a clear understanding of the body's rotational behavior.

Moment of Inertia

The moment of inertia is a pivotal concept in rotational dynamics, reflecting how a rigid body's mass is distributed relative to an axis of rotation. It governs the angular acceleration that the body will experience under applied torque. The greater the moment of inertia, the more resistant the body is to changes in rotational velocity.

For a point mass, the moment of inertia is simply the product of the mass \(m\) and the square of its perpendicular distance \(r\) from the axis of rotation:
\[I = m r^2\]
However, for a rigid body, we must integrate this quantity over the entire volume of the body to consider the mass distribution. The tasks of finding eigenvalues and identifying the principal axis of a rigid body are intimately connected to its moments of inertia. Each principal axis has a corresponding eigenvalue, which is the body's moment of inertia about that axis.

When we dive into the inertia tensor calculation for any point on the principal axis, as shown in the exercise, we solidify our understanding that the moment of inertia, as an eigenvalue, remains unchanged. This concept reassures us of the stability of rotational motion around the principal axes and underscores the integral role the moment of inertia plays in dynamics.