Chapter 2: Problem 38
Show that the center of mass is well defined, i.e., does not depend on the choice of the origin of reference for radius vectors.
Chapter 2: Problem 38
Show that the center of mass is well defined, i.e., does not depend on the choice of the origin of reference for radius vectors.
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Get started for freeFor which values of \(\alpha\) is motion along a circular orbit in the field with potential energy \(U=r^{\alpha},-2 \leq \alpha<\infty\), Liapunov stable?
Show that if a field is axially symmetric and conservative, then its potential energy has the form \(U=U(r, z)\), where \(r, \varphi\), and \(z\) are cylindrical coordinates. In particular, it follows from this that the vectors of the ficld lie in planes through the \(z\) axis.
Show that the set of phase curves on the surface \(\pi_{E_{0}}\) forms a twodimensional sphere. The formula \(w=\left(x_{1}+i y_{1}\right) /\left(x_{2}+i y_{2}\right)\) gives the "Hopf map" from the three sphere \(\pi_{E_{0}}\) to the two sphere (the complex \(w\)-plane completed by the point at infinity). Our phase curves are the pre-images of points under the Hopf map.
Is the field in the plane minus the origin given by \(F_{1}=x_{2} /\left(x_{1}^{2}+x_{2}^{2}\right)\), \(F_{2}=-x_{1} /\left(x_{1}^{2}+x_{2}^{2}\right)\) conservative? Show that a field is conservative if and only if its work along any closed contour is equal to zero.
Draw the image of the circle \(x^{2}+(y-1)^{2}<\frac{1}{4}\) under the action of a transformation of the phase flow for the equations (a) of the "inverse pendulum," \(\ddot{x}=x\) and \((\) b \()\) of the "nonlinear pendulum," \(\ddot{x}=-\sin x\).
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