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A \(2.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((-1.20 \mathrm{~m}, 0.500 \mathrm{~m})\) and a \(4.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((0.600 \mathrm{~m},-0.750 \mathrm{~m})\) Both lie on a horizontal plane. At what (a) \(x\) and (b) \(y\) coordinates must you place a \(3.00 \mathrm{~kg}\) particle such that the center of mass of the three-particle system has the coordinates \((-0.500 \mathrm{~m},-0.700 \mathrm{~m}) ?\)

Short Answer

Expert verified
Place the particle at coordinates (-1.50 m, -2.10 m).

Step by step solution

01

Determine the Formula for Center of Mass

The center of mass (COM) for multiple particles in the x-direction is given by the formula: \[ x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} \] and similarly, for the y-direction: \[ y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} \]. Here, \( m_i \) are the masses and \( (x_i, y_i) \) are the coordinates of the particles.
02

Set Up the Equation for x-coordinate

We know the center of mass's x-coordinate is \(-0.500\ m\). Plug in the given values: \( m_1 = 2.00 \ kg, x_1 = -1.20 \ m; m_2 = 4.00 \ kg, x_2 = 0.600 \ m; m_3 = 3.00 \ kg, x_3 = x \); \(-0.500\ m = \frac{2.00 \times (-1.20) + 4.00 \times 0.600 + 3.00 \times x}{2.00 + 4.00 + 3.00} \).
03

Solve the x-direction Equation

Calculate inside the fraction: \(-2.40 + 2.40 + 3x = -0.500 \times 9.00\), resulting in \(3x = -4.50\). Solving for \(x\), we find \(x = \frac{-4.50}{3}\), giving \(x = -1.50 \ m\).
04

Set Up the Equation for y-coordinate

For the y-coordinate, the center of mass is \(-0.700 \ m\); use the equation with known values: \\(-0.700 = \frac{2.00 \times 0.500 + 4.00 \times (-0.750) + 3.00 \times y}{2.00 + 4.00 + 3.00} \).
05

Solve the y-direction Equation

Calculate inside the fraction: \(1.00 - 3.00 + 3y = -0.700 \times 9.00\), resulting in \(3y = -6.30\). Solving for \(y\), we find \(y = \frac{-6.30}{3}\), giving \(y = -2.10 \ m\).

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Key Concepts

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

Particle System
In physics, a **particle system** refers to a collection of particles that interact with each other and possibly with external forces. For this exercise, we are looking at a system of three particles, each with a specific mass and position in a two-dimensional plane. When analyzing particle systems, especially for calculating the center of mass, it's crucial to consider each particle's contribution based on its mass and its coordinates.

The center of mass for a system of particles is a point that represents the average position of the entire mass of the system. This is particularly useful in understanding how the system behaves as a whole, instead of focusing on individual particles. Calculating the center of mass involves taking a weighted average of the positions of particles; here, the weight is the mass of each particle.

For students, it's important to recognize that setting up the problem requires careful consideration of all included particles and making sure coordinates are assigned properly. With this understanding, you can move forward to accurately solve the problem.
Coordinate Calculation
When tasked with calculating the center of mass of a particle system, understanding **coordinate calculation** is essential. This involves using the formula for the center of mass separately for the x and y coordinates. This can be expressed as:
  • direction: \[ x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} \]
  • direction: \[ y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} \]
Here, each term in the numerator of the formula represents one particle's contribution, considering both its mass and position.

In our solution, these calculations require substituting the known values into the formulas. By solving these equations, you determine where an additional particle needs to be placed, to result in a specific center of mass for the entire system. Remember that when coordinates are negative, they should be treated as such, impacting how they influence the center of mass.
Mass Distribution
In this context, **mass distribution** refers to how the mass is spread across different particles in the system. Each particle's mass contributes to where the center of mass of the system is located, making it a crucial factor in solving these types of problems.

The particles involved here have varying masses — 2.00 kg, 4.00 kg, and 3.00 kg. A heavier particle will generally have a larger impact on the system's center of mass because the center is a weighted average of all masses. Thus, balancing these masses accurately determines the resultant center.

While calculating, recognizing the role mass plays allows us to balance the equations correctly, ensuring that the center of mass reflects the system's physical configuration. If you see how these distributions shift the center, visualizing and solving the problem becomes easier and more intuitive. Understanding mass distribution helps in predicting how adding or shifting mass in the system will alter the center of mass.

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Most popular questions from this chapter

In a common but dangcrous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\). and the collision on the floor lasts \(0.082 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

A stone is dropped at \(t=0 .\) A second stone, with twice the mass of the first, is dropped from the same point at \(t=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stoncs at \(t=300 \mathrm{~ms}^{7}\) (Ncither stonc has yct reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

A \(2140 \mathrm{~kg}\) railroad flatcar, which can move with negligible friction, is motionless next to a platform. A \(242 \mathrm{~kg}\) sumo wrestler runs at \(5.3 \mathrm{~m} / \mathrm{s}\) along the platform (parallel to the track) and then jumps onto the flatcar. What is the speed of the flatcar if he then (a) stands on it, (b) runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to it in his original direction, and (c) turns and runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to the flatcar opposite his original dircction?

an \(80 \mathrm{~kg}\) man is on a ladder hanging from a balloon that has a total mass of \(320 \mathrm{~kg}\) (including the basket passenger). The balloon is initially stationary relative to the ground. If the man on the ladder begins to climb at \(2.5 \mathrm{~m} / \mathrm{s}\) relative to the ladder, (a) in what direction and (b) at what speed does the balloon move? (c) If the man then stops climbing, what is the speed of the balloon?

A rocket is moving away from the solar system at a speed of \(6.0 \times 10^{3} \mathrm{~m} / \mathrm{s} .\) It fires its engine, which ejects exhaust with a speed of \(3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to the rocket. The mass of the rocket at this time is \(4.0 \times 10^{4} \mathrm{~kg},\) and its acceleration is \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the thrust of the engine? (b) At what rate, in kilograms per second, is exhaust ejected during the firing?

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