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Chapter 1: Problem 6

A mechanical system consists of three points. At the initial moment their velocities (in some inertial coordinate system) are equal to zero. Show that the points always remain in the plane which contained them at the initial moment.

Short Answer

Expert verified
Question: Show that three points with initial velocities equal to zero in a mechanical system will always remain in the plane which contained them at the initial moment. Answer: By calculating the change in the normal vector of the plane containing the points A, B, and C, we proved that the time derivative of the normal vector is equal to zero. This means that the normal vector does not change over time, and therefore, the points A, B, and C always remain in the plane which contained them at the initial moment.

Step by step solution

01

Define the position vectors of the points

Assume that the initial positions of the three points are A, B, and C. Define the position vectors of these points as \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) respectively.
02

Calculate the relative position vectors

Find the relative position vectors between the points, which are the difference of the initial position vectors: \(\vec{AB} = \vec{b} - \vec{a}\) \(\vec{AC} = \vec{c} - \vec{a}\)
03

Calculate the normal vector of the initial plane

To define a plane that contains the three points, we can take the cross product of the relative position vectors, which will give us the normal vector of the plane: \(\vec{n} = \vec{AB} \times \vec{AC}\)
04

Find the velocities of the points

Since the initial velocities of all points are zero, their velocities at any later time depend on the forces acting on them, which we'll assume are conservative. Without loss of generality, we can write the velocity vectors of points A, B, and C as: \(\vec{v_A} = k_A \cdot \vec{F_A}\) \(\vec{v_B} = k_B \cdot \vec{F_B}\) \(\vec{v_C} = k_C \cdot \vec{F_C}\) where \(k_A\), \(k_B\), and \(k_C\) are scalar constants and \(\vec{F_A}\), \(\vec{F_B}\), and \(\vec{F_C}\) are the force vectors at points A, B, and C, respectively.
05

Calculate the change in relative position vectors

As time progresses, the relative position vectors between points will change. To find this change, take the time derivative of \(\vec{AB}\) and \(\vec{AC}\): \(\frac{d\vec{AB}}{dt} = \frac{d(\vec{b} - \vec{a})}{dt} = \vec{v_B} - \vec{v_A}\) \(\frac{d\vec{AC}}{dt} = \frac{d(\vec{c} - \vec{a})}{dt} = \vec{v_C} - \vec{v_A}\)
06

Calculate change in normal vector

The normal vector will also change with time. Take the time derivative of \(\vec{n}\) using the product rule: \(\frac{d\vec{n}}{dt} = (\vec{v_B} - \vec{v_A}) \times \vec{AC} + \vec{AB} \times (\vec{v_C} - \vec{v_A})\)
07

Show that normal vector change is zero

To prove that the points remain in the initial plane, we need to show that the change in the normal vector is equal to zero. Substitute the expressions of the velocities for points A, B, and C, and expand the cross products in the previous step: \(\frac{d\vec{n}}{dt} = (k_B \cdot \vec{F_B} - k_A \cdot \vec{F_A}) \times \vec{AC} + \vec{AB} \times (k_C \cdot \vec{F_C} - k_A \cdot \vec{F_A})\) Now, we'll use the property of the cross product that states \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\): \(\frac{d\vec{n}}{dt} = (k_B \cdot \vec{F_B} \times \vec{AC} - k_A \cdot \vec{F_A} \times \vec{AC}) + (\vec{AB} \times k_C \cdot \vec{F_C} - \vec{AB} \times k_A \cdot \vec{F_A})\) Conservative forces can be represented as the negative gradient of a potential function, i.e., \(\vec{F} = -\nabla U\). The cross product of the force vector with the relative position vector is then zero: \(\vec{F} \times \vec{R} = -(\nabla U) \times \vec{R} = 0\) Thus, each term in the expression for \(\frac{d\vec{n}}{dt}\) is zero, and we can conclude that the change in the normal vector is zero: \(\frac{d\vec{n}}{dt} = 0\) Therefore, the points A, B, and C always remain in the plane which contained them at the initial moment.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inertial Coordinate System
Understanding the concept of an inertial coordinate system is foundational in the realms of classical mechanics. An inertial frame of reference is essentially a description of space where Newton's laws of motion take their simplest form. This means that in such a system, a body with no forces acting on it moves with constant velocity—which could very well be zero, implying a state of rest.

In the context of the problem from the textbook, the velocities of the points being zero initially tell us that we are working within an inertial frame. The absence of external influences means that unless a force acts on these points, they will remain stationary or maintain constant velocity. It’s like throwing a ball in a spaceship far away from gravity's pull—it will keep moving at a consistent speed and direction unless you interfere. That baseline behavior is crucial for predicting the motion and interactions of objects, helping us to deduce that if the points started in a plane, they will continue to move in that same plane unless acted upon by a force.
Velocity Vectors
Velocity vectors are arrows pointing in the direction an object is moving, with a length proportional to the speed; these are mathematical entities that provide tremendous insight into motion dynamics. In the problem we’re examining, we initially have the points at rest, which means their velocity vectors are all zero. As time passes and forces start to act – assuming they are conservative – the velocity of each point will change in a way that’s both predictable and directly related to those forces.

When we think of these vectors, it's like visualizing an arrow shooting from each point, dictating exactly how fast and in which direction the point should travel. As the exercise unfolds, considering the vectors as results of conservative forces emphasizes the interconnected nature of velocity and force. The velocity vectors reflect how each point gets buffeted by its surroundings, influenced by the potential energy within the system.
Conservative Forces
Conservative forces are a class of forces that have a crucial property: the work done by these forces is independent of the path taken. What that boils down to is the fact that these forces conserve mechanical energy within a system. The classic examples of conservative forces include gravity and electrostatic forces – think of a roller coaster looping around tracks, no matter the path it returns to the same point, its potential energy in relation to the Earth's gravity is the same.

In our textbook problem, the conservative nature of the forces means that we can represent them as a gradient of a potential energy function. Since the forces can be expressed this way, when they are crossed with a relative position vector, they consistently lead to zero. This mathematical nugget is pivotal for the final step in our problem: it’s the reason why the normal vector to the plane containing the points doesn’t change with time, thus preserving the motion of points within the initial plane. In summary, it’s as though these points are tethered to an invisible sheet that guides their dance, ensuring they never stray from their two-dimensional stage.

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Most popular questions from this chapter

Show that every gaillean transformation of the space \(\mathbb{R} \times \mathbb{R}^{3}\) can be written in a unique way as the composition of a rotation, a translation, and a uniform motion \(\left(g=g_{1} \circ g_{2} \circ g_{3}\right)\) (thus the dimension of the galilean group is equal to \(3+4+3=10\) ).

Show that all galilean spaces are isomorphic to each other \({ }^{6}\) and, in particular, isomorphic to the coordinate space \(\mathbb{R} \times \mathbb{R}^{3}\). Let \(M\) be a set. A one-to-one correspondence \(\varphi_{1}: M \rightarrow \mathbb{R} \times \mathbb{R}^{3}\) is called a galilean coordinate system on the set \(M\). A coordinate system \(\varphi_{2}\) moves uniformly with respect to \(\varphi_{1}\) if \(\varphi_{1} \varphi_{2}^{-1}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R} \times \mathbb{R}^{3}\) is a galilean transformation. The galilean coordinate systems \(\varphi_{1}\) and \(\varphi_{2}\) give \(M\) the same galilean structure.

A mechanical system consists of two points. Show that for any initial conditions there exists an inertial coordinate system in which the two points remain in a fixed plane.

Show that if a mechanical system consists of only one point, then its acceleration in an inertial coordinate system is equal to zero ("Newton's first law").

A mechanical system consists of two points. At the initial moment their velocities (in some inertial coordinate system) are equal to zero. Show that the points will stay on the line which connected them at the initial moment.
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