Textbook Question
A thin, 100 g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the speed of a point on the rim?
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Ch 12: Rotation of a Rigid Body
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 12, Problem 13a
The four masses shown in FIGURE EX12.13 are connected by massless, rigid rods. Find the coordinates of the center of mass.
Verified step by step guidance1
Step 1: Understand the problem. The center of mass (COM) is the weighted average position of all the masses in the system. To find the coordinates of the center of mass, we use the formulas: \( x_{COM} = \frac{\sum m_i x_i}{\sum m_i} \) and \( y_{COM} = \frac{\sum m_i y_i}{\sum m_i} \), where \( m_i \) is the mass of each object, and \( x_i \) and \( y_i \) are the coordinates of each mass.
Step 2: Extract the data from the image. The masses and their coordinates are: \( m_A = 70 \, \text{g}, (x_A, y_A) = (0, 0) \), \( m_B = 190 \, \text{g}, (x_B, y_B) = (20, 0) \), and \( m_C = 400 \, \text{g}, (x_C, y_C) = (20, 12) \).
Step 3: Calculate the total mass \( \sum m_i \). Add the masses: \( \sum m_i = m_A + m_B + m_C \).
Step 4: Compute \( x_{COM} \). Use the formula \( x_{COM} = \frac{m_A x_A + m_B x_B + m_C x_C}{\sum m_i} \). Substitute the values of \( m_i \) and \( x_i \) from the data.
Step 5: Compute \( y_{COM} \). Use the formula \( y_{COM} = \frac{m_A y_A + m_B y_B + m_C y_C}{\sum m_i} \). Substitute the values of \( m_i \) and \( y_i \) from the data.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Center of Mass
The center of mass of a system is the point where the total mass of the system can be considered to be concentrated. It is calculated as the weighted average of the positions of all masses in the system, taking into account their respective masses. The formula for the center of mass in two dimensions is given by the coordinates (x_cm, y_cm) = (Σ(m_i * x_i) / Σm_i, Σ(m_i * y_i) / Σm_i), where m_i is the mass and (x_i, y_i) are the coordinates of each mass.
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Intro to Center of Mass
Mass Distribution
Mass distribution refers to how mass is spread out in a system. In this problem, the masses A, B, and C are located at specific coordinates, and their distribution affects the calculation of the center of mass. Understanding how different masses contribute to the overall center of mass is crucial, as heavier masses will have a greater influence on the position of the center of mass compared to lighter ones.
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Coordinate System
A coordinate system provides a framework for locating points in space using numerical values. In this problem, a Cartesian coordinate system is used, where each point is defined by its x (horizontal) and y (vertical) coordinates. This system is essential for calculating the center of mass, as it allows for precise determination of the positions of the masses involved in the calculation.
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Related Practice
Textbook Question
The three 200 g masses in FIGURE EX12.11 are connected by massless, rigid rods. What is the triangle's kinetic energy if it rotates about the axis at 5.0 rev/s?
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Textbook Question
What is the rotational kinetic energy of the earth? Assume the earth is a uniform sphere. Data for the earth can be found inside the back cover of the book.
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Textbook Question
The three masses shown in FIGURE EX12.15 are connected by massless, rigid rods. Find the coordinates of the center of mass.
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Textbook Question
The four masses shown in FIGURE EX12.13 are connected by massless, rigid rods. Find the moment of inertia about a diagonal axis that passes through masses B and D.
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Textbook Question
The three masses shown in FIGURE EX12.15 are connected by massless, rigid rods. Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.
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