The result (10.7), that E mar = 0, can be paraphrased to say that the position vector of the CM relative to the CM is zero, and, in this form, is nearly obvious. Nevertheless, to be sure you understand the result, prove it by solving (10.4) for ea and substituting into the sum concerned.
Read moreTable of Contents
Textbook Solutions for Classical Mechanics
Question
Consider an arbitrary rigid body with an axis of rotational symmetry, which we'll call \(\hat{\mathbf{z}}\).
(a) Prove that the axis of symmetry is a principal axis.
(b) Prove that any two directions \(\hat{\mathbf{x}}\) and \(\hat{\mathbf{y}}\) perpendicular to \(\hat{\mathbf{z}}\) and each other are also principal axes.
(c) Prove that the principal moments corresponding to these two axes are equal: \(\lambda_{1}=\lambda_{2}\).
Solution
Step 1 of 3
(a) If the axis is an axis of rotational symmetry, then we saw in Example 10.1(c) that \(I_{x z}=I_{y z}=0\).
Therefore, If \(\omega=(0,0, \omega)\), then \(\mathbf{L}=\left(I_{x z} \omega, I_{y z} \omega, I_{z z} \omega\right)=\left(0,0, I_{z z} \omega\right)\) and \(\hat{\mathbf{Z}}\) is a principal axis.
full solution
Watch The Answer!
Principal Axes and Moments in Rotational Symmetry
Want To Learn More? To watch the entire video and ALL of the videos in the series:
full solution
Unravel the mysteries of rotational symmetry and principal axes in this video. We prove the relationships between the axis of symmetry and principal axes, showcasing the equal principal moments. Join us for an exploration of the fundamental concepts of rigid body dynamics.