/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Two particles, each of mass \(2.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two particles, each of mass \(2.90 \times 10^{-4} \mathrm{~kg}\) and speed \(5.46 \mathrm{~m} / \mathrm{s}\), travel in opposite directions along parallel lines separated by \(4.20 \mathrm{~cm} .\) (a) What is the magnitude \(L\) of the angular momentum of the two-particle system around a point midway between the two lines? (b) Does the value of \(L\) change if the point about which it is calculated is not midway between the lines? If the direction of travel for one of the particles is reversed, what would be (c) the answer to part (a) and (d) the answer to part (b)?

Short Answer

Expert verified
(a) \(6.64308 \times 10^{-6} \mathrm{~kg{m}^2{/}s} \), (b) No, (c) Zero, (d) No change.

Step by step solution

01

Identify the formulas

To calculate the angular momentum, we can use the formula for a point mass: \( L = mvr \), where \( m \) is the mass, \( v \) is the speed, and \( r \) is the perpendicular distance to the point about which the angular momentum is calculated.
02

Calculate individual angular momenta

Since the two particles travel in opposite directions, their angular momenta around the midpoint will be in the same direction (additive). Each particle has a distance \( r = 2.10 \mathrm{~cm} = 0.021 \mathrm{~m} \) from the midpoint. Thus, the angular momentum for each particle is \( L = (2.90 \times 10^{-4} \mathrm{~kg})(5.46 \mathrm{~m/s})(0.021 \mathrm{~m}) \).
03

Compute individual angular momentum

Substitute the values into the formula to compute the angular momentum for one particle: \( L = 2.90 \times 10^{-4} \times 5.46 \times 0.021 = 3.32154 \times 10^{-6} \mathrm{~kg{m}^2{/}s} \).
04

Total angular momentum about the midpoint

Since the angular momenta of the two particles add up, the total angular momentum \( L_{total} \) is twice that of a single particle: \( L_{total} = 2 \times 3.32154 \times 10^{-6} = 6.64308 \times 10^{-6} \mathrm{~kg{m}^2{/}s} \).
05

Analyze dependence on calculation point (Part b)

The total angular momentum would not change based on the calculation point as long as the same distance from each particle line is maintained, ensuring symmetrical contributions.
06

Effect of reversing one particle's direction (Part c)

If one particle reverses direction, their angular momenta will be opposite and subtract from each other so that \( L_{total} = 0 \).
07

Analysis of point independence in reversed case (Part d)

Even if not midway, with one particle reversed, the net angular momentum remains zero, as opposite contributions cancel regardless of point placement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Two-Particle System
In physics, a two-particle system often involves considering two separate masses or particles that may interact or have a specific configuration relative to one another. Understanding how systems with more than one particle behave is essential for solving problems related to momentum and angular momentum.

When dealing with angular momentum, the choice of calculation may significantly depend on the relative positions of the particles and their velocities. In a two-particle system, each particle contributes to the total angular momentum.
  • When particles travel in opposite directions, their momenta can either add up or cancel out, depending on the point of calculation.
  • In this specific exercise, while traveling along parallel lines, each particle affects the total angular momentum based on its perpendicular distance to the chosen axis (or point of calculation).
For a complete understanding, you should visualize that each particle is treated individually, then combine their influences to find the system's total angular momentum.
Parallel Lines
Parallel lines, in a geometrical context, are distinct lines running alongside one another at a consistent distance apart and never intersecting. In the case of angular momentum for two particles moving along parallel lines, these line details can significantly influence calculations.

In this exercise, parallel lines ensure that both particles are moving the same perpendicular distance from a point, simplifying the angular momentum calculation:
  • Both lines are equidistant throughout their length, maintaining consistent perpendicular distances from any selected midpoint.
  • This is crucial because the angular momentum calculation depends on these distances remaining constant.
Understanding the nature of parallel lines helps when reasoning why the angular momentum is additive or subtractive, depending on the direction of particle travel.
Point Mass
A point mass is a simplified model in physics that represents an object's mass concentration at a single point in space for analytical convenience. This simplification is particularly useful in angular momentum problems to ease calculations related to different axes and points of rotation.

In our exercise scenario:
  • Each particle is treated as a point mass, which means its mass is concentrated at a single point along its path.
  • This simplification allows for straightforward use of the formula for angular momentum: \( L = mvr \).
  • Here, \( m \) is the mass of each particle, \( v \) is the velocity, and \( r \) is the perpendicular distance — all focusing on the point mass concept.
By considering particles as point masses, we can directly apply this methodology to find individual and total angular momenta quickly and accurately.
Perpendicular Distance
Perpendicular distance is the shortest distance between a point and a specific line. In angular momentum calculations, the perpendicular distance from the point of calculation to where particles travel is integral to determining the magnitude of momentum. When considering angular momentum:
  • The distance perpendicular to the path of the particle is used because it is the shortest possible distance, making it the most accurate when calculating momentum based on circular motion.
  • In this exercise, this distance was 2.10 cm (converted to meters), representing half the separation between the two parallel paths of the particles.
  • Any change in where you measure from (while maintaining this perpendicular distance) doesn't alter calculated angular momentum — ensuring accuracy when using centralized points like above two parallel lines.
Understanding and maintaining perpendicular distances allow for precise angular momentum calculations across various physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of \(0.10 g ?\) (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to \(0.10 g\) ? Why?

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\). The rotational inertia of the record about its axis of rotation is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2} .\) A wad of wet putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800\) \(\mathrm{m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2} .\) By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

The uniform rod (length \(0.60\) \(\mathrm{m}\), mass \(1.0 \mathrm{~kg}\) ) in Fig. \(11-54\) rotates in the plane of the figure about an axis through one end, with a rotational inertia of \(0.12 \mathrm{~kg} \cdot \mathrm{m}^{2}\). As the rod swings through its lowest position, it collides with a \(0.20 \mathrm{~kg}\) putty wad that sticks to the end of the rod. If the rod's angular speed just before collision is \(2.4 \mathrm{rad} / \mathrm{s}\), what is the angular speed of the rod-putty system immediately after collision?

A wheel of radius \(0.250 \mathrm{~m}\), which is moving initially at \(43.0\) \(\mathrm{m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of \((\mathrm{a})\) its lin- ear acceleration and (b) its angular acceleration. (c) The wheel's rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. Calculate the magnitude of the torque about the central axis due to friction on the wheel.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.