/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 \(\cdot\) Three odd-shaped block... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\cdot\) Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: \((1) \quad 0.300\) kg, \((0.200 \mathrm{m}, 0.300 \mathrm{m}) ;\) (2) \(0.400 \mathrm{kg}, \quad(0.100 \mathrm{m},-0.400 \mathrm{m})\) (3) \(0.200 \mathrm{kg},(-0.300 \mathrm{m}, 0.600 \mathrm{m}) .\) 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 approximately (0.044 m, 0.056 m).

Step by step solution

01

Calculate the Total Mass

To find the total mass of the system, add together the masses of all three blocks. \[ m_{\text{total}} = 0.300\, \text{kg} + 0.400\, \text{kg} + 0.200\, \text{kg} = 0.900\, \text{kg} \]
02

Calculate the X-coordinate of the Center of Mass

The x-coordinate of the center of mass \((x_{\text{cm}})\) is calculated using the formula: \[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_{\text{total}}} \] Substitute the given values: \[ x_{\text{cm}} = \frac{(0.300)(0.200) + (0.400)(0.100) + (0.200)(-0.300)}{0.900} \] Calculate each term: \[ (0.300 \times 0.200) = 0.060, \ (0.400 \times 0.100) = 0.040, \ (0.200 \times -0.300) = -0.060 \] Sum these up: \[ 0.060 + 0.040 - 0.060 = 0.040 \] Divide by the total mass: \[ x_{\text{cm}} = \frac{0.040}{0.900} \approx 0.044 \text{ m} \]
03

Calculate the Y-coordinate of the Center of Mass

The y-coordinate of the center of mass \((y_{\text{cm}})\) is calculated similarly: \[ y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_{\text{total}}} \] Substitute the given values: \[ y_{\text{cm}} = \frac{(0.300)(0.300) + (0.400)(-0.400) + (0.200)(0.600)}{0.900} \] Calculate each term: \[ (0.300 \times 0.300) = 0.090, \ (0.400 \times -0.400) = -0.160, \ (0.200 \times 0.600) = 0.120 \] Sum these up: \[ 0.090 - 0.160 + 0.120 = 0.050 \] Divide by the total mass: \[ y_{\text{cm}} = \frac{0.050}{0.900} \approx 0.056 \text{ m} \]
04

Conclusion

Combine the calculated x and y coordinates to obtain the center of mass of the system: \((0.044\, \text{m}, 0.056\, \text{m})\). This is the center of mass of all three chocolate blocks combined.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mass Distribution
Understanding mass distribution is crucial when calculating the center of mass. Imagine several objects, each having its own unique mass and position. These masses must be "distributed" across a certain space. In our chocolate block example, each block has a different mass and position. The center of mass is influenced by both the individual mass of each object and how these masses are placed or distributed in space.

Mass distribution involves knowing:
  • Each object's mass
  • The position of each mass within the chosen coordinate system
In short, by considering how the masses are spread out, we can determine the point that represents the average position — the center of mass. This helps in understanding the balance and movement of the combined system.
Coordinate System
To find the center of mass, it is essential to use a clear and defined coordinate system. A coordinate system helps in pinpointing the precise location of each mass point in space. Let's look at some essentials of using a coordinate system:
  • The coordinate system includes axes, usually labeled as x, y, and sometimes z for three-dimensional problems.
  • Each block of chocolate in our problem is positioned at a specific point on this system, denoted as \((x, y)\) coordinates.
By assigning each mass a coordinate position, you can easily plug these values into formulas to find the center of mass. Think of it like plotting points on a map — without a coordinate system, calculating the center of mass would be nearly impossible!
Calculation Steps
Calculating the center of mass involves straightforward math once you understand the distribution and coordinate system. Here's a streamlined view of the steps involved:
  • First, determine the total mass of the system by summing up the masses of all individual objects.
  • Next, calculate the weighted average for the x-coordinates. This involves multiplying each mass by its x-position and summing these products.
  • Divide this sum by the total mass to find the x-coordinate of the center of mass.
  • Repeat the process for the y-coordinate using the y-positions of each mass.
Remember, it’s like blending colors. Each mass contributes to the final position statement — the combined point that represents the entire system best: the center of mass.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A catcher catches a 145 g baseball traveling horizontally at 36.0 \(\mathrm{m} / \mathrm{s}\) . (a) How large an impulse does the ball give to the catcher? (b) If the ball takes 20 \(\mathrm{ms}\) to stop once it is in contact with the catcher's glove, what average force did the ball exert on the catcher?

\(\bullet\) Detecting planets around other stars. Roughly 500 planets have so far been detected beyond our solar system. This is accomplished by looking for the effect the planet has on the star. The star is not truly stationary; instead, it and its planets orbit around the center of mass of the system. Astronomers can measure this wobble in the position of a star.(a) For a star with the mass and size of our sun and having a planet with five times the mass of Jupiter, where would the center of mass of this system be located, relative to the center of the star, if the distance from the star to the planet was the same as the distance from Jupiter to our sun? (Consult Appendix E.) (b) If the planet had earth's mass, where would the center of mass of the system be located if the planet was just as far from the star as the earth is from the sun? (c) In view of your results in parts (a) and (b), why is it much easier to detect stars having large planets rather than small ones?

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 \(\mathrm{kg}\) and ejects gas at a relative velocity of magnitude 2000 \(\mathrm{m} / \mathrm{s}\) , how much gas must it eject in the first second to have an initial acceleration of 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) .

Two identical 1.50 \(\mathrm{kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{N} / \mathrm{cm},\) compress- ing the spring by 20.0 \(\mathrm{cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal lab table.

\(\bullet\) A 20.0 -kg lead sphere is hanging from a hook by a thin wire 3.50 m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00 -kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a com- plete circular loop after the collision?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.