Lec11(2) - Finding Moments of Inertia 1 2.003J/1.053J Dynamics and Control I Spring 2007 Professor Thomas Peacock Lecture 11 2D Motion of Rigid Bo

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Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 1 Finding Moments of Inertia 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 3/14/2007 Lecture 11 2D Motion of Rigid Bodies: Finding Moments of Inertia, Rolling Cylinder with Hole Example Finding Moments of Inertia Figure 1: Rigid Body. Figure by MIT OCW. I C = m i | ρ i | 2 i = m i ( x 2 i + y i 2 ) i I C is the Moment of Inertia about C.
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Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. ± 2 Finding Moments of Inertia Example: Uniform Thin Rod of Length L and Mass M Figure 2: Uniform thin rod of length L and mass M . Figure by MIT OCW. I C = m i ( x i 2 + y i 2 )For very thin rod, y i is small enough to neglect. i m i x i 2 i Rod has mass/length = ρ . Convert to integral. I C = x 2 dm rod dm = ρdx ± L/ 2 I C = x 2 ρdx L/ 2 ² 3 ³ L/ 2 L 3 x = ρ = ρ 3 12 L/ 2 We know that mass M = ρL . Therefore: ML 2 I C = 12
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Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT
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This note was uploaded on 11/29/2011 for the course CIVIL 1.018j taught by Professor Markusbuehler during the Fall '08 term at MIT.

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Lec11(2) - Finding Moments of Inertia 1 2.003J/1.053J Dynamics and Control I Spring 2007 Professor Thomas Peacock Lecture 11 2D Motion of Rigid Bo

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