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IA University of South Alabama MA227 y WileyPLUS Ahmad > c Secure l https://edugen.wileyplus.com/edugen/student/mainfr.uni Bing w Wikipedia R Facebook y Twitter-LinkedIn D The weather Channel Yelp »Other Bookmarks I Help I Contact Us I Leg Out ps WileyPLUS Differential calculus. LA University of South D Apple iCloud Yahoo WileyPLUS Halliday, Fundamentals of Physics, 10e Calculus-based Physics I ⅈ (PH 201-202) Home Read, Study & Practice Gradebook ORION Downloadable eTextbook Assianment> Open Assignment FULL SCREEN PRINTER VERSION BACK ASSIGNMENT RESOURCES Chapter 10, Problem 038 201-14 The figure shows three 0.0134 kg particles that have been glued to a rod of length L-6.41 cm and negligible mass. The assembly can rotate around a perpendicular axis through point o at the left end. If we remove one particle (that is, 33% of the mass), by what percentage does the rotational inertia of the assembly around the rotation axis decrease when that removed particle is (a) the innermost one and (b) the outermost one? 034 Axis 0 048 050 054 -+-+ (a) Number Units (b) Number Units LINK TO SAMPLE PROBLEM LINK TO SAMPLE ROBLEM Question Attempts: 0 of 6 used SAVE FOR LATER SUBMIT ANSWER Version 4.24.1.23

Explanation / Answer

I = m R^2

In this case, the inertia of the system of three particles would be:

I = md^2 + m(2d)^2 + m(3d)^2 = 14 md^2

a)When we remove the innermost one, Inertia of the system decreases and becomes:

I = m(2d)^2 + m (3d)^2 = 13 md^2

So, %age decrease would be:

%age = (14 - 13)/14 x 100 = 7.14 %

Hence, when the innermost mass is removed, inertia decreases by 7.14 %

b)When the outer most is remover, the inertia of system becomes:

I = md^2 + m(2d)^2 = 5 md^2

So, the %age decrease would be:

%age = (14 - 5 )/14 x 100 = 64.3 %

Hence, when the outermost mass is removed, the inertia decreases by 64.3%

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