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Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). Show that the angular momentum of this system of particles is the same about any point taken as the origin.

Short Answer

Expert verified
The system's angular momentum is \( mdv \) about any origin.

Step by step solution

01

Understanding Angular Momentum

Angular momentum for a particle can be defined as the cross product of its position vector \( \mathbf{r} \) and its linear momentum \( \mathbf{p} \). For the system of two particles, we have:\[ \mathbf{L} = \mathbf{r}_1 \times \mathbf{p}_1 + \mathbf{r}_2 \times \mathbf{p}_2, \]where \( \mathbf{p}_i = m\mathbf{v}_i \) is the momentum of the \( i^{th} \) particle.
02

Position Vectors and Moment Calculation

Consider a coordinate system where the two parallel lines run along the x-axis separated by distance \( d \). Let particle 1 be at position \( (x_1, d/2) \) and particle 2 at \( (x_2, -d/2) \), both moving along the x-axis with velocities \( v \) and \( -v \) respectively.
03

Calculating Angular Momentum of Each Particle

For particle 1:- Position vector \( \mathbf{r}_1 = (x_1, d/2) \).- Momentum \( \mathbf{p}_1 = m(v, 0) \).- Angular momentum: \( \mathbf{L}_1 = \mathbf{r}_1 \times \mathbf{p}_1 = m[x_1(v)(0) - (d/2)(0)] = 0 \).For particle 2:- Position vector \( \mathbf{r}_2 = (x_2, -d/2) \).- Momentum \( \mathbf{p}_2 = m(-v, 0) \).- Angular momentum: \( \mathbf{L}_2 = \mathbf{r}_2 \times \mathbf{p}_2 = m[x_2(-v)(0) - (-d/2)(0)] = mdv \).
04

Total Angular Momentum

The total angular momentum is given by:\[ \mathbf{L}_{total} = \mathbf{L}_1 + \mathbf{L}_2 = 0 + mdv = mdv \].This calculation demonstrates that the angular momentum due to particle rotation about any point in the plane is \( mdv \, \text{(counter-clockwise)} \).
05

Consideration of Any Origin

Since the result of the cross products only involves the perpendicular distance \( d \) and constant speed \( v \), moving the origin to any other point (sideways along the y-axis or otherwise) won't affect this angular momentum magnitude due to the parallel nature of the motion of both particles.

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Key Concepts

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

Cross Product
The cross product is vital in calculating angular momentum. When dealing with vectors, such as the position vector \( \mathbf{r} \) and linear momentum \( \mathbf{p} \), the cross product measures the perpendicular impact of one vector on another. For angular momentum, the cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) describes how the momentum impacts a rotating system. \( \mathbf{L} \) results in a new vector that is perpendicular to both \( \mathbf{r} \) and \( \mathbf{p} \). This characteristic is crucial because the magnitude of this vector, \( |\mathbf{L}| = |\mathbf{r}||\mathbf{p}|\sin(\theta) \), depends directly on the angle \( \theta \) between \( \mathbf{r} \) and \( \mathbf{p} \). In the case of parallel lines, though the directional effect is simplified, the perpendicular distance from the line of action remains, affirming angular momentum. Understanding and applying the cross product helps in various vector operations, especially in physics.
Linear Momentum
Linear momentum is a measure of an object's motion, given by the product of its mass and velocity. It is a fundamental value used to understand how particles move and interact. In our example, each particle's linear momentum is calculated as \( \mathbf{p} = m\mathbf{v} \), where \( m \) stands for mass and \( \mathbf{v} \) represents velocity. This simple relationship allows us to predict the effects of different forces and how they will influence a particle’s motion. - Linear momentum is always directed along the path of motion.- Conserves in a closed system without external forces.- Forms the basis for concepts like impulse and collisions.When particles in the example travel along parallel lines with opposing velocities, their linear momentums contribute uniquely to the system's angular momentum.
Position Vector
A position vector represents a point's location in space relative to a defined origin. It helps to determine other vector quantities, like angular momentum. In two dimensions, a position vector \( \mathbf{r} \) can be described with coordinates \((x, y)\), expressing the point's displacement from an origin. When particles move along parallel lines, the position vector becomes crucial as it determines the perpendicular distance contributing to angular momentum.- Position vector specifies a particle's current position.- Connects an object’s path with origin-defined coordinates.- Essential for calculating translational and rotational movements.In our exercise, the position vectors \( \mathbf{r}_1 = (x_1, d/2) \) and \( \mathbf{r}_2 = (x_2, -d/2) \) help establish the basis from which angular momentum can be computed straightforwardly.
Parallel Lines
Parallel lines are lines in a plane that never intersect. They maintain a constant distance apart, which has unique implications in physics, especially concerning motion and forces. - Parallel lines ensure uniformity in certain motion properties.- In physics, make comparing traversing moving particles easier.- Influence how forces and resulting momenta function relative to one another.In the given exercise, both particles travel along parallel lines, separated by a fixed distance \( d \). This arrangement contributes to the visibility of the consistent angular momentum across various observation points. Because of their parallel nature, regardless of the origin's selection, the angular momentum around any point in the system is preserved, owing to the equal and opposite linear momenta relative to each line's separation.

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Most popular questions from this chapter

A solid ball of radius \(6.0 \mathrm{~cm}\) is initially rolling smoothly at 10 \(\mathrm{m} / \mathrm{s}\) along a horizontal floor. It then rolls smoothly up a ramp until it momentarily stops. What maximum height above the floor does it reach?

Two disks are mounted (like a merry-go-round) on lowfriction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia \(3.30 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at \(450 \mathrm{rev} / \mathrm{min}\). The second disk, with rotational inertia \(6.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at \(900 \mathrm{rev} / \mathrm{min}\). They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at \(900 \mathrm{rev} / \mathrm{min}\), what are their (b) angular speed and (c) direction of rotation after they couple together?

In unit vector notation, find the net torque about the origin on a particle located at \((2.0 \mathrm{~m},-2.0 \mathrm{~m},-6.0 \mathrm{~m})\) when the three forces \(\vec{F}_{1}=(6.0 \mathrm{~N}) \hat{\mathrm{j}}, \vec{F}_{2}^{t}=(1.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}\), and \(\vec{F}_{3}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+\) \((2.0 \mathrm{~N}) \hat{\mathrm{j}}-(3.0 \mathrm{~N}) \mathrm{k}\) act on the particle.

A top spins at \(25 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.2 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

At the instant the displacement of a \(1.50 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

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