Problem 77 - Fundamentals of Physics

Question

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) Is the value different for a different location of the point? If the direction of either particle is reversed, what are the answers for (c) part (a) and (d) part (b)?

Step-by-Step Solution

Answer

(a) \(6.64 \times 10^{-5} \mathrm{~kg} \cdot \mathrm{m}^2/\mathrm{s}\); (b) No; (c) 0; (d) No.

Step 1: Understanding Angular Momentum

Angular momentum for a particle moving in a straight line can be calculated as \( L = r \times p \), where \( r \) is the perpendicular distance to the line of motion (moment arm) and \( p \) is the linear momentum given by \( p = mv \). The total angular momentum is the sum of the angular momentum of each particle.

Step 2: Calculate Linear Momentum

Each particle has a mass \( m = 2.90 \times 10^{-4} \mathrm{~kg} \) and speed \( v = 5.46\mathrm{~m} / \mathrm{s} \). Thus, the linear momentum of one particle is \( p = m \cdot v = (2.90 \times 10^{-4}) \times 5.46 \approx 1.58 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m/s} \).

Step 3: Determine Perpendicular Distance

The perpendicular distance from the midpoint between the lines to the line of motion for each particle is half of the separation distance, which is \( rac{4.20 \mathrm{~cm}}{2} = 2.10 \mathrm{~cm} = 0.021 \mathrm{~m} \).

Step 4: Calculate Angular Momentum for Each Particle

The angular momentum of each particle can be calculated as \( L = r \times p = 0.021 \times 1.58 \times 10^{-3} \approx 3.32 \times 10^{-5} \mathrm{~kg} \cdot \mathrm{m}^2/\mathrm{s} \).

Step 5: Total Angular Momentum (Part a)

Since they are moving in opposite directions, the angular momentum contributions add up. Therefore, the total angular momentum is \( 2 \times 3.32 \times 10^{-5} = 6.64 \times 10^{-5} \mathrm{~kg} \cdot \mathrm{m}^2/\mathrm{s} \).

Step 6: Location Dependency (Part b)

Since the system's center of mass is the same and the perpendicular distance to the center stays unchanged, the value of the angular momentum is independent of the exact location, as long as the point is equidistant from the two lines.

Step 7: Reversed Direction Impact (Parts c and d)

If the direction of one particle is reversed, the momenta will cancel each other out, leading to zero angular momentum. Thus, for part (c), the angular momentum is 0. For part (d), the location of the point still does not affect the zero total angular momentum.

Key Concepts

Linear Momentum

Linear momentum is a fundamental quantity in physics that describes the motion of an object. It is a vector, meaning it has both magnitude and direction. The formula to calculate linear momentum (\( p \)) is:

• \( p = mv \)

where:

• \( m \) is the mass of the object.
• \( v \) is the velocity of the object.

In this exercise, each particle has a mass \( m = 2.90 \times 10^{-4} \mathrm{~kg} \) and moves at a speed \( v = 5.46 \mathrm{~m/s} \).
This results in a linear momentum of approximately \( 1.58 \times 10^{-3} \mathrm{~kg \cdot m/s} \) per particle. Since they move in opposite directions, each has the same magnitude of linear momentum, but one has a positive direction and the other a negative.
This property is critical in determining the overall angular momentum of the system.

Moment Arm

The concept of a moment arm is essential when calculating angular momentum because it gives the perpendicular distance from the reference point to the line of action of a force. In this case, the moment arm refers to the perpendicular distance from a point (midway between the two lines) to the path of the particle.

• The exercise states this distance is half of the separation between the lines, thus \( \frac{4.20 \mathrm{~cm}}{2} = 2.10 \mathrm{~cm} \) or \( 0.021 \mathrm{~m} \).

The moment arm acts like a leverage factor, amplifying the effect of the force applied. Its role in angular momentum can be explained by the equation:

• \( L = r \times p \)

where:

• \( L \) is the angular momentum.
• \( r \) is the moment arm.
• \( p \) is the linear momentum.

In this scenario, the moment arm determines how much each particle's linear momentum contributes to the total angular momentum.

Center of Mass

Understanding the center of mass is crucial for analyzing the motion of a system of particles. It represents a specific point where the entire mass of the system can be considered to be concentrated. In this problem, the center of mass lies equidistant from the motion lines of both particles.

• Since the particles move in a symmetric fashion, the center of mass remains at the midpoint between the two lines, simplifying the calculation of angular momentum.

The stability in the position of the center of mass, relative to the chosen point, means that the angular momentum calculation remains consistent across different scenarios.
In essence, even if the reference point changes location but stays equidistant from the lines, the calculated angular momentum will remain the same until a particle's direction is reversed.
This consistency is important as it affects the overall dynamics and ensures the system behaves predictably under symmetrical conditions.