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Chapter 6: Problem 6

Find the configuration space of a rigid body, all of whose points lie on a line. }

Short Answer

Expert verified
Answer: The configuration space of a rigid body, all of whose points lie on a line, is a one-dimensional space consisting of its position along the line.

Step by step solution

01

Determine the degrees of freedom

A rigid body can have multiple degrees of freedom in general, based on the specific constraints given. For a one-dimensional rigid body that only moves along a line, there are only two possible degrees of freedom: translation (along the line) and rotation (around the line).
02

Definition of translation and rotation

Let's define the translation and rotation more precisely. Translation refers to the position of the body along the line (measured using a coordinate along the line). Rotation refers to the orientation (angle), which denotes how the rigid body has rotated around the line, measured using an angle with respect to a reference point. Note that the rotation degree of freedom is irrelevant since the orientation of the rigid body does not change when it rotates around its axis.
03

Configuration space

Since only the translation is relevant here, the configuration space is one-dimensional, and given directly by the position along the line. Let x represent the position of the rigid body along the line, then the configuration space q can be written as: q = x So the configuration space of a rigid body, all of whose points lie on a line, is a one-dimensional space consisting of its position along the line.

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Most popular questions from this chapter

Show that the moments of inertia of any body satisfy the triangle inequalities $$ I_{3} \leq I_{2}+I_{1} \quad I_{2} \leq I_{1}+I_{3} \text { and } I_{1} \leq I_{2}+I_{3} \text {. } $$ and that equality holds only for a planar body.

Find the principal axes and moments of inertia of a uniform tetrahedron relative to its vertices.

A small mass \(\varepsilon\) is added to a rigid body with moments of inertia \(I_{1}>I_{2}>I_{3}\) at the point \(\mathbf{Q}=x_{1} \mathbf{e}_{1}+x_{2} \mathbf{e}_{2}+x_{3} \mathbf{e}_{3}\). Find the change in \(I_{1}\) and \(\mathbf{e}_{1}\) with error \(\mathrm{O}\left(\varepsilon^{2}\right)\). Solution. The center of mass is displaced by a distance of order \(\varepsilon\). Therefore, the moments of inertia of the old body with sespect to the parallel axes passing through the old and new centers of mass differ in magnitude of order \(\varepsilon^{2}\). At the same time, the addition of mass changes the moment of inertia relative to any fixed axis by order \(\varepsilon\). Therefore, we can disregard the displacement of the center of mass for calculations with error \(\mathrm{O}\left(\varepsilon^{2}\right)\). Thus, after addition of a small mass the kinetic energy takes the form $$ T=T_{0}+\frac{1}{2} \varepsilon[\Omega, Q]^{2}+O\left(\varepsilon^{2}\right) $$ where \(T_{0}=\frac{1}{2}\left(I_{1} \Omega_{1}^{3}+I_{2} \Omega_{2}^{2}+I_{3} \Omega_{3}^{2}\right)\) is the kinetic energy of the original body. We look for the eigenvalue \(I_{t}(\varepsilon)\) and eigenvector \(\mathbf{e}_{1}(\varepsilon)\) of the inertia operator in the form of a Taylor series in \(\varepsilon\). By equating coefficients of \(\varepsilon\) in the relation \(A(\varepsilon) \mathbf{e}_{1}(\varepsilon)=I_{1}(\varepsilon) e_{1}(\varepsilon)\), we find that, within error \(\mathrm{O}\left(\varepsilon^{2}\right)\) : $$ I_{1}(\varepsilon) \approx I_{1}+\alpha\left(x_{2}^{2}+x_{3}^{2}\right) \text { and } \mathbf{e}_{1}(\varepsilon) \approx \mathbf{e}_{1}+\varepsilon\left(\frac{x_{1} x_{2}}{I_{2}-I_{1}} \mathbf{e}_{2}+\frac{x_{1} x_{3}}{I_{3}-I_{1}} \mathbf{e}_{3}\right) $$ From the formula for \(I_{1}(\varepsilon)\) it is clear that the change in the principal moments of inertia (to the first approximation in \(\varepsilon\) ) is as if neither the center of mass nor the principal axes changed. The formula for \(e_{1}(\varepsilon)\) demonstrates how the directions of the principal axes change: the largest principal axis of the inertia ellipsoid approaches the added point, and the smallest recedes from it. Furthermore, the addition of a small mass on one of the principal planes of the inertia ellipsoid rotates the two axes lying in this plane and does not change the direction of the third axis. The appearance of the differences of moments of inertia in the denominator is connected with the fact that the major axes of an ellipsoid of revolution are not defined. If the inertia ellipsoid is nearly an ellipsoid of revolution (i.e., \(I_{1} \approx I_{2}\) ) then the addition of a small mass could strongly turn the axes \(e_{1}\) and \(e_{2}\) in the plane spanned by them.

A river flows with velocity \(3 \mathrm{~km} / \mathrm{hr}\). For what radius of curvature of a river bend is the Coriolis force from the earth's rotation greater than the centrifugal force determined by the flow

Are stationary rotations of the body around the largest and smallest principal axes Liapunov stable?
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