Textbook Question
A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door’s moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge?
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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 16a
A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door's moment of inertia for rotation on its hinges?
Verified step by step guidance1
Step 1: Understand the problem. The moment of inertia (I) is a measure of an object's resistance to rotational motion about a specific axis. For this problem, the door is treated as a rectangular object rotating about one of its edges (the hinges). The formula for the moment of inertia of a rectangular object about an edge parallel to one of its sides is: , where m is the mass and w is the width of the door.
Step 2: Convert the dimensions of the door into meters, as SI units are required for consistency. The width of the door is given as 91 cm, which is equivalent to 0.91 m. The height of the door (220 cm) is not relevant for this calculation since the rotation is about the hinges along the height.
Step 3: Substitute the given values into the formula. The mass of the door (m) is 25 kg, and the width (w) is 0.91 m. The formula becomes: .
Step 4: Simplify the expression by squaring the width (0.91 m) and multiplying it by the mass (25 kg). Then, multiply the result by 1/3 to find the moment of inertia.
Step 5: The final result will give the moment of inertia of the door for rotation about its hinges. Ensure the units are in kg·m², as this is the standard unit for moment of inertia.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. For a solid rectangular object, the moment of inertia can be calculated using specific formulas that take into account its dimensions and mass.
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Intro to Moment of Inertia
Axis of Rotation
The axis of rotation is the line about which an object rotates. In this case, the door rotates around its hinges, which serve as the axis. The position of this axis significantly affects the moment of inertia, as the distribution of mass relative to the axis determines how difficult it is to rotate the object.
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13:46Parallel Axis Theorem
Parallel Axis Theorem
The parallel axis theorem allows us to calculate the moment of inertia of an object about any axis parallel to an axis through its center of mass. It states that the moment of inertia about the new axis is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes. This theorem is useful when dealing with objects that do not rotate about their center of mass.
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Related Practice
Textbook Question
A 12-cm-diameter DVD has a mass of 21 g. What is the DVD’s moment of inertia for rotation about a perpendicular axis through its center?
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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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Textbook Question
A 12-cm-diameter DVD has a mass of 21 g. What is the DVD's moment of inertia for rotation about a perpendicular axis through the edge of the disk?
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