HW 9b bonus HW, not requried for exams

# HW 9b bonus HW, not requried for exams - HW#9b Page 1 of 6...

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HW#9b Page 1 of 6 Note: numbers used in solution steps are different from your WebAssign values. For a hoop all mass are distributed at distance “R” away from the center. So rotating along the center axis perpendicular to the hoop, I =mR 2 , regardless of the thickness, because mass distribution along the axis doesn’t change I, as long as mass are not distributed closer or further from the rotation axis. For a cylinder most mass are distributed closer to the center than R, respect to its center axis, I =mR 2 , regardless of thickness.

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HW#9b Page 2 of 6 Values I’ll use: 1 kg, 3 kg at either end of a 3 m rod. Makes sense. Center of mass is closer to larger mass. Specifically, since m 2 is double m 1 (2kg vs. 1kg). Note: This is not a bar of uniform mass (do not use I=1/3ML 2 ). This is a massless bar with a point mass on each end You should simply treat the 2 mass as points at the end and sum mr 2 for the two mass points. Notice that r, the distance from the mass to the rotation axis changes when you change axis.
HW#9b Page 3 of 6

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HW#9b Page 4 of 6 Values I’ll use: Each panel has m=10kg F=50N Approach 1: Treat as four panels spun through edge, each of m=10kg and L=4m Approach 2: Treat as two doors spun through center, each of m=20kg and L=8 m Note: both approaches yield same value for I. H is irrelevant in both approaches. Using either approach is fine. Force is applied to only one panel.
HW#9b Page 5 of 6 Values I’ll use: m A =3kg, m B =5kg Center of mass at the 47cm mark of meter stick.

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## This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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HW 9b bonus HW, not requried for exams - HW#9b Page 1 of 6...

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