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Chapter 8: Problem 51

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: (1) 0.300 kg, (0.200 m, 0.300 m); (2) 0.400 kg, (0.100 m, \(-\)0.400 m); (3) 0.200 kg, (\(-\)0.300 m, 0.600 m). Find the coordinates of the center of mass of the system of three chocolate blocks.

Short Answer

Expert verified
The center of mass is at (0.044 m, 0.056 m).

Step by step solution

01

Identify given data

We are given the masses and center-of-mass coordinates for three blocks of chocolate: - Block 1: Mass = 0.300 kg, Coordinates = (0.200 m, 0.300 m) - Block 2: Mass = 0.400 kg, Coordinates = (0.100 m, -0.400 m) - Block 3: Mass = 0.200 kg, Coordinates = (-0.300 m, 0.600 m)
02

Use the center of mass formula

The center of mass (CM) of a system is given by the formula: \[ x_{CM} = \frac{\sum{m_i x_i}}{\sum{m_i}} \] and \[ y_{CM} = \frac{\sum{m_i y_i}}{\sum{m_i}} \] where \(m_i\) are the masses and \((x_i, y_i)\) are the coordinates of each mass.
03

Calculate the total mass

Calculate the total mass of the blocks:\[ m_{\text{total}} = 0.300 \, \text{kg} + 0.400 \, \text{kg} + 0.200 \, \text{kg} = 0.900 \, \text{kg} \]
04

Calculate x-coordinate of CM

Calculate the x-coordinate of the center of mass:\[ x_{CM} = \frac{(0.300 \times 0.200) + (0.400 \times 0.100) + (0.200 \times (-0.300))}{0.900} \]\[ = \frac{0.060 + 0.040 - 0.060}{0.900} \]\[ = \frac{0.040}{0.900} \approx 0.044 \text{ m} \]
05

Calculate y-coordinate of CM

Calculate the y-coordinate of the center of mass:\[ y_{CM} = \frac{(0.300 \times 0.300) + (0.400 \times (-0.400)) + (0.200 \times 0.600)}{0.900} \]\[ = \frac{0.090 - 0.160 + 0.120}{0.900} \]\[ = \frac{0.050}{0.900} \approx 0.056 \text{ m} \]
06

Compile the CM coordinates

The coordinates for the center of mass of the system are approximately (0.044 m, 0.056 m).

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Key Concepts

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

Understanding Mass Distribution
When solving problems related to the center of mass, understanding mass distribution is crucial. Mass distribution refers to how mass is spread out in space. In the exercise, each chocolate block has a different mass and position, which contributes differently to the system's center of mass.
  • Each block's mass acts as a weight at its respective coordinates.
  • Heavier blocks have a larger impact on the overall center of mass location.
The way mass is distributed affects the balance of the system. Think of it like balancing a seesaw; where the mass is located determines the balance point. Properly calculating this distribution allows us to pinpoint the exact center. In our exercise, calculating the mass distribution involves multiplying each block's mass by its respective coordinates. This provides insight into how each block contributes to the total system's center of mass.
Exploring Coordinate Systems
Coordinate systems are essential in problems involving positions, like center of mass calculations. In our exercise, each block's position is given in a Cartesian coordinate system, typically using an (x, y) format. This allows you to visualize each block's location in space.
  • The coordinates indicate where on the plane each block is positioned.
  • An (x, y) value of (0, 0) would be the origin or the central point of the plane.
  • Positive and negative values determine directions along the axes.
With these coordinates, we calculate an average or weighted balance of the system components through the formula for the center of mass. Understanding the coordinate system is imperative because errors in interpreting these coordinates can lead to incorrect conclusions about the center of mass.
Introduction to Composite System Analysis
Composite system analysis involves understanding how multiple objects, or blocks in our exercise, interact to form a larger system.
  • A composite system combines individual systems or objects into one whole.
  • The behavior and properties of the system depend on the interaction between its components.
In our example, analyzing the composite system means considering each block's mass and position to find the collective center of mass. The center of mass represents the "average" location of all the mass distribution. This analysis allows us to understand how changing one component might affect the overall system. This can be particularly useful in engineering, physics, and design where balancing components precisely is crucial to functionality and safety.

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Most popular questions from this chapter

A ball with mass \(M\), moving horizontally at 4.00 m/s, collides elastically with a block with mass 3\(M\) that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit.

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 6.50-kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has 1.75 kg of water in its cavity, at what speed must it expel this water suddenly to achieve a speed of 2.50 m/s to escape the predator? Ignore any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20.0 kg. Burt has mass 30.0 kg; Ernie has mass 40.0 kg. Initially, the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Ignore any horizontal force that the water exerts on the log, and assume that neither friend falls off the log.

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. (a) A \(25.0-\mathrm{kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity. (b) A 25.0 -kg mass is thrown back ward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A 25.0 -kg mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750-kg car traveling to the right at 1.50 m/s collides with a 1450-kg car going to the left at 1.10 m/s. Measurements show that the heavier car's speed just after the collision was 0.250 m/s in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.
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