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Four objects are situated along the \(y\) axis as follows: a \(2.00 \mathrm{kg}\) object is at \(+3.00 \mathrm{m},\) a \(3.00-\mathrm{kg}\) object is at \(+2.50 \mathrm{m}\) a \(2.50-\mathrm{kg}\) object is at the origin, and a \(4.00-\mathrm{kg}\) object is at \(-0.500 \mathrm{m} .\) Where is the center of mass of these objects?

Short Answer

Expert verified
The Center of mass of the objects is at \(0.611 m\) along the \(y\) axis.

Step by step solution

01

Identification of given parameters

Identify the given masses and their respective positions. We have four objects with masses \(m_1 = 2.00 kg\) at position \(y_1 = +3.00 m\), \(m_2 = 3.00 kg\) at position \(y_2 = +2.50 m\), \(m_3 = 2.50 kg\) at position \(y_3 = 0 m\), and \(m_4 = 4.00 kg\) at position \(y_4 = -0.500 m\).
02

Apply the formula for center of mass

The formula for the center of mass is given by \(y_{cm} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i}\), where \(m_i\) and \(y_i\) are the masses and positions of the objects. Apply this formula using the given values.
03

Calculation of the numerator

Multiplicity of each mass with its respective position to calculate the numerator part of the formula: \((2.00 kg * +3.00 m) + (3.00 kg * +2.50 m) + (2.50 kg * 0 m) + (4.00 kg * -0.500 m)\)
04

Calculate the denominator

Add up all the given masses to calculate the denominator part of the formula: \(2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg\)
05

Divide the numerator by the denominator

Take the result from step 3 and divide it by the result from step 4 to find the center of mass.

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Key Concepts

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

Mass Distribution
In physics, understanding how mass is distributed across objects is crucial for various calculations, especially when determining the center of mass. Mass distribution refers to the way mass is spread out in space, influencing the overall balance of an object or system.
Consider four objects positioned at different points along the vertical line. Each object has a specific mass and location on the y-axis. This arrangement directly impacts how the mass is distributed.
  • The total mass consists of the individual masses of all the objects;
  • The position and mass of each object affect their contribution to the center of mass;
  • Identifying the precise location of each object is crucial to understand mass distribution.
By examining their individual positions and masses, you can visualize how the mass is spread along the y-axis, forming a continuous distribution. This visualization helps in predicting how forces, like gravity, will interact with each object differently based on its mass and position.
Coordinate System
A coordinate system is essential for describing the position of objects in space. Here, we focus on a one-dimensional coordinate system along the y-axis, simplifying the task of locating objects. The y-axis is a straight line where you assign numerical values to different points to communicate their position clearly.
In our scenario, all objects are placed on the y-axis at specific points, either positive or negative. This linear approach allows you to:
  • Easily reference positions based on distance from a chosen origin;
  • Use positive or negative values to indicate direction along the axis;
  • Work with simple numerical calculations to determine positional relationships.
The y-axis is a fundamental tool for organizing data in one dimension. It enables a straightforward mathematical approach suitable for calculating the center of mass by providing clarity on where each object lies in relation to others.
Mathematical Calculation
Mathematical calculations play a central role in determining the center of mass. The center of mass is a point where you can effectively consider the entire mass of a system to be concentrated.
The formula to find the center of mass along a single axis is:\[ y_{cm} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i} \]This formula involves:
  • Calculating the product of each mass and its respective position;
  • Adding these products to form the numerator;
  • Summing up all masses to form the denominator;
  • Finally, dividing the total of the numerator by the total mass in the denominator to find the center of mass value.
By substituting the specific values from our example, you methodically determine where the center of mass lies on the y-axis. Each step is crucial in forming a complete understanding of how positions and weights collectively influence this pivotal point.

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Most popular questions from this chapter

A 2.00 -kg particle has a velocity \((2.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s},\) and a \(3.00-\mathrm{kg}\) particle has a velocity \((1.00 \hat{\mathbf{i}}+6.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) Find (a) the velocity of the center of mass and (b) the total momentum of the system.

The first stage of a Saturn \(V\) space vehicle consumed fuel and oxidizer at the rate of \(1.50 \times 10^{4} \mathrm{kg} / \mathrm{s}\), with an exhaust speed of \(2.60 \times 10^{3} \mathrm{m} / \mathrm{s} .\) (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle's initial mass was \(3.00 \times 10^{6} \mathrm{kg} .\) Note: You must include the gravitational force to solve part (b).

A \(90.0-\mathrm{kg}\) fullback running east with a speed of \(5.00 \mathrm{m} / \mathrm{s}\) is tackled by a \(95.0-\mathrm{kg}\) opponent running north with a speed of \(3.00 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly inelastic, (a) calculate the speed and direction of the players just after the tackle and (b) determine the mechanical energy lost as a result of the collision. Account for the missing energy.

An orbiting spacecraft is described not as a "zero-g," but rather as a "microgravity" environment for its occupants and for on-board experiments. Astronauts experience slight lurches due to the motions of equipment and other astronauts, and due to venting of materials from the craft. Assume that a 3500 -kg spacecraft undergoes an acceleration of \(2.50 \mu g=2.45 \times 10^{-5} \mathrm{m} / \mathrm{s}^{2}\) due to a leak from one of its hydraulic control systems. The fluid is known to escape with a speed of \(70.0 \mathrm{m} / \mathrm{s}\) into the vacuum of space. How much fluid will be lost in 1 h if the leak is not stopped?

The mass of the Earth is \(5.98 \times 10^{24} \mathrm{kg},\) and the mass of the Moon is \(7.36 \times 10^{22} \mathrm{kg} .\) The distance of separation, measured between their centers, is \(3.84 \times 10^{8} \mathrm{m} .\) Locate the center of mass of the Earth-Moon system as measured from the center of the Earth.

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