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Chapter 4: Problem 5

An astronaut is floating in empty space at rest relative to an inertial frame with her arms by her side. Explain how it is that by waving her arms and then returning them to their original position, she can rotate her body, but cannot move her centre of mass.

Short Answer

Expert verified
Answer: An astronaut can rotate her body in space without changing her center of mass by waving her arms and returning them to the original position. This action causes changes in angular momentum due to internal forces, while the conservation of angular momentum principle ensures the total angular momentum remains constant. Since no external forces are acting on the astronaut, the position of her center of mass remains unchanged relative to the inertial frame.

Step by step solution

01

1. Understanding Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts upon a system, then the total angular momentum of the system remains constant. In this case, the astronaut is floating in space with no external forces acting on her, therefore her total angular momentum remains constant.
02

2. Understanding Center of Mass

The center of mass is a hypothetical point of a system where the entire mass of that system is concentrated. It is the average position of all the mass particles that make up the system. The center of mass of a system only moves when it is acted upon by an external force. As mentioned before, no external forces are acting on the astronaut, so her center of mass should remain constant in position.
03

3. Analyzing the Astronaut's Motion

Initially, the astronaut is at rest, which means her initial angular momentum is zero. As she starts to wave her arms, her body parts experience internal forces. However, the total angular momentum of her body must remain constant as per the conservation principle mentioned above. Therefore, if her arms gain some angular momentum in a certain direction, her body must gain an equal amount of angular momentum in the opposite direction to maintain the total angular momentum.
04

4. Returning Arms to Original Position

When the astronaut returns her arms to their original position, she effectively gives her body's angular momentum back to her arms. The internal torques within her body result in a rotation with a total angular momentum equal to the negative of that when her arms were extended. This change in momentum causes her body to rotate.
05

5. Center of Mass Remains Unchanged

Throughout the entire motion, no external forces act upon the astronaut. The changes in her body's configuration (i.e., waving and returning her arms) only lead to changes in the position of her body parts while leaving the center of mass unchanged. This is because the motion of her arms is governed by internal forces, and they do not affect the position of her center of mass relative to the inertial frame. In conclusion, the astronaut can rotate her body by waving her arms and returning them to the original position due to the conservation of angular momentum. However, since no external forces are acting on her and the motion of her arms involves solely internal forces, the position of her center of mass remains unchanged relative to the inertial frame.

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Most popular questions from this chapter

The configuration space for a particle of mass \(m\) moving in space is Euclidean space, with Cartesian coordinates \(x, y, z\). Show that the generator of the one-parameter group of rotations about the \(x\)-axis is $$ u=(0,-z, y) \text {. } $$ Hence write down an expression for the \(x\)-component of angular momentum in terms of the spherical polar coordinates defined by $$ \begin{aligned} &x=r \sin \theta \cos \varphi \\ &y=r \sin \theta \sin \varphi \\ &z=r \cos \theta . \end{aligned} $$

Let \(\rho_{s}: C \times \mathbb{R} \rightarrow C \times \mathbb{R}\) be a dynamical symmetry of a system with Lagrangian \(L\), with generators \(u_{a}\). Show that under the action of \(\rho_{s}\) for small \(s\), the change in \(\partial L / \partial v_{a}\) along a kinematic trajectory is $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=s u_{b} \frac{\partial^{2} L}{\partial v_{a} \partial q_{b}}+s w_{b} \frac{\partial^{2} L}{\partial v_{a} \partial v_{b}} $$ to the first order in \(s\), where $$ w_{a}=\frac{\partial u_{a}}{\partial q_{b}} v_{b}+\frac{\partial u_{a}}{\partial t} $$ Deduce that $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ and $$ \delta\left(\frac{\partial L}{\partial q_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial q_{b}}-s \frac{\partial w_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ Show that $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial u_{b}}{\partial q_{a}}\right)=\frac{\partial w_{b}}{\partial q_{a}} $$ Hence show that if \(q_{a}=q_{a}(t)\) is a solution of Lagrange's equations, then so is \(q_{a}=q_{a}(t)+s u_{a}(q(t), t)\), to the first order in \(s\). Deduce that if \(\rho_{s}\) is a dynamical symmetry, then it maps dynamical trajectories in \(C \times \mathbb{R}\) to dynamical trajectories.

A particle of unit mass moves under gravity on a smooth surface given in cylindrical polar coordinates \(z, r, \theta\) by \(z=f(r)\). Show that the motion is governed by the Lagrangian $$ L=\frac{1}{2} \dot{r}^{2}\left(1+f^{\prime}(r)^{2}\right)+\frac{1}{2} r^{2} \dot{\theta}^{2}-g f(r) . $$ Show that \(\theta\) is an ignorable coordinate. Write down the conserved conjugate momentum and give its physical interpretation.

A system of particles with masses \(m_{\alpha}\) moves in a gravitational field \(\boldsymbol{g}\). Show that if \(\boldsymbol{F}_{\alpha}=-m_{\alpha} \boldsymbol{g}\) where \(\boldsymbol{g}\) is constant, then \(\sum_{\alpha} \boldsymbol{F}_{\alpha}=m \boldsymbol{g} \quad\) and \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}=m \boldsymbol{c} \wedge \boldsymbol{g}\), where \(c\) is the position vector of the centre of mass and \(m\) is the total mass. Deduce that the effect of a uniform gravitational fieldon the total linear momentum and angular momentum is the same as that of a single force \(m g\) acting through the centre of mass. Show by counter-example that this is not true for a non-uniform field.
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