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 5
Two balls are connected by a 150-cm-long massless rod. The center of mass is 35 cm from a 75 g ball on one end. What is the mass attached to the other end?
Verified step by step guidance1
Step 1: Understand the concept of the center of mass. The center of mass is the point where the weighted relative position of the masses is balanced. For a system of two masses connected by a rod, the center of mass can be calculated using the formula: , where is the center of mass position, and are the masses, and and are their respective distances from a reference point.
Step 2: Assign the given values. The length of the rod is 150 cm, so the distance between the two masses is 150 cm. The center of mass is located 35 cm from the 75 g ball. Let the mass of the other ball be , and its distance from the center of mass will be cm.
Step 3: Write the center of mass equation for this system. Using the reference point at the position of the 75 g ball, the equation becomes: . Here, the position of the 75 g ball is taken as 0 cm.
Step 4: Rearrange the equation to solve for . Multiply through by the denominator to eliminate the fraction: . Expand and simplify: .
Step 5: Isolate by moving terms involving to one side: . Simplify further: . Finally, solve for by dividing both sides by 80: .
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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. In this problem, the center of mass is used to determine the balance point between the two balls connected by the rod.
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Intro to Center of Mass
Mass and Weight
Mass is a measure of the amount of matter in an object, typically measured in grams or kilograms, while weight is the force exerted by gravity on that mass. In this scenario, the mass of the second ball needs to be calculated based on the position of the center of mass and the known mass of the first ball, illustrating the relationship between mass distribution and gravitational force.
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Equilibrium and Torque
In a system in equilibrium, the sum of torques around any point must be zero. This principle can be applied to the two balls connected by the rod, where the torque produced by the weight of each ball about the center of mass must balance out. Understanding this concept is essential for solving the problem, as it allows for the calculation of the unknown mass based on the known distances and mass.
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Related Practice
Textbook Question
A ceiling fan with 80-cm-diameter blades is turning at 60 rpm. Suppose the fan coasts to a stop 25 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off?
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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
A ceiling fan with 80-cm-diameter blades is turning at 60 rpm. Suppose the fan coasts to a stop 25 s after being turned off. Through how many revolutions does the fan turn while stopping?
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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
An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60 rpm to 90 rpm in 10 s. What is the tangential acceleration of a tooth on the sprocket?
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