For a thin, circular ring, what is its mass moment of inertia? Is it Ixx = Iyy = (mr^2)/4, and Izz = (mr^2)/2? Also, if I have integral(dm), for this ring will dm = (m/(2*pi*r)) * (r*dtheta). Let me know if I am wrong

For a thin, circular ring, what is its mass moment of inertia? Is it Ixx = Iyy = (mr^2)/4, and Izz = (mr^2)/2? Also, if I have integral(dm), for this ring will dm = (m/(2pir)) * (r*dtheta). Let me know if I am wrong

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.8P: Using integration, compute the polar moment of inertia about point O for the circular sector. Check...

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Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provide resistance in changing its state of velocity or motion for coming to a state of rest. In other words, Newton's first law of motion describes the law of inertia of a rigid body.

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For a thin, circular ring, what is its mass moment of inertia? Is it Ixx = Iyy = (mr^2)/4, and Izz = (mr^2)/2?

Also, if I have integral(dm), for this ring will dm = (m/(2*pi*r)) * (r*dtheta). Let me know if I am wrong.

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