Problem 74 - College Physics

Question

(a) The center of mass of a system consisting of two 0.10-kg particles is located at the origin. If one of the particles is at \((0,0.45 \mathrm{~m}),\) where is the other? \((\mathrm{b})\) If the masses are moved so their center of mass is located at \((0.25 \mathrm{~m}\), \(0.15 \mathrm{~m}\) ), can you tell where the particles are located?

Step-by-Step Solution

Answer

(a) The other particle is at \((0, -0.45 \text{ m})\). (b) No, you cannot uniquely determine their positions.

Step 1: Understanding the Center of Mass Formula

For a system of particles, the center of mass (COM) is calculated using the formula \( \vec{R}_{cm} = \frac{1}{M} \sum_i m_i \vec{r}_i \), where \( M \) is the total mass of the system, \( m_i \) is the mass of each particle, and \( \vec{r}_i \) is the position vector of each particle. For a two-particle system, this simplifies to \( \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \).

Step 2: Applying the Formula for Part (a)

Given that the center of mass is at the origin \((0,0)\) and one particle is at \((0,0.45 \text{ m})\), we set the equation \( 0 = \frac{0.10 \times (0,0.45) + 0.10 \times \vec{r}_2}{0.10 + 0.10} \). Solve for \( \vec{r}_2 \): \( 0 = 0.05 \times (0,0.45) + 0.05 \times \vec{r}_2 \), thus \( \vec{r}_2 = (0, -0.45) \).

Step 3: Understanding Part (b)

In part (b), the new center of mass position is given as \((0.25 \text{ m}, 0.15 \text{ m})\). With two equal masses and a known center of mass, the formula does not provide unique positions for the particles. The arrangement of particles is dependent on their relative distances from this new center of mass.

Step 4: Analyzing Unique Determination of Positions

Since the masses are identical and the COM is now at a specific point, there are infinite possibilities for positions of the individual particles that could result in the described COM. The relative positions of particles can vary as long as their combined mass maintains the COM at \((0.25 \text{ m}, 0.15 \text{ m})\).

Key Concepts

Particle System

In physics, a particle system consists of a collection of particles, each of which can be described by its mass and position. Consider a simple system with two particles. Each particle has its mass and a point in space where it is located. These basic elements can define complex behaviors when interacting within a larger system.

• Each particle has a defined position, described by coordinates in a space (e.g., 2D or 3D).
• The behavior of the entire system can be analyzed through the combined properties of its particles, such as total mass and center of mass.
• In applying real-world scenarios, particle systems can model anything from celestial bodies to atomic structures.

Understanding the particle system concept is crucial because it allows us to analyze how individual particles influence the system as a whole.

Mass Distribution

Mass distribution refers to how mass is spread out across the particles in a system. It affects how the system behaves and moves.
For a two-particle system, each particle contributes to the total distribution of mass based on its location and mass.

• The location of mass affects the system's center of mass (COM), which is the average location of all mass in the system.
• Even if masses are equal, as in our exercise, the distribution is essential for determining the COM position.
• Changes in the positions of particles alter the COM, which can influence the system's stability and dynamics.

This concept helps us understand translating particle properties into larger system dynamics.

Position Vector

A position vector expresses the location of a point in space relative to an origin. It's vital in determining the center of mass in a system of particles.
For each particle, its position vector is an essential component of calculating where the COM lies.

• Position vectors are generally expressed in coordinate form, such as \(x, y, z\) for three dimensions.
• For a two-particle system, knowing one particle's position vector allows you to calculate another if the COM is known.
• The formula \(\vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}\) shows how position vectors are used to find the COM.

Understanding position vectors grants insight into particle arrangement in any space.

Equal Masses

When particles in a system have equal masses, some calculations simplify; however, certain outcomes may become ambiguous, as seen in part (b) of the exercise. Equal masses allow for the variables of distance and positioning to become more influential on the system's behavior.

• Both particles contribute equally to the center of mass, meaning their positioning dramatically affects the outcome.
• As seen in this exercise, with equal masses and a known center of mass, there are infinite possible positions for the particles.
• This property can cause ambiguity, as exact positions are not easily determinable without additional constraints.

Such concepts highlight why equal masses are significant in computations of dynamics and stability.