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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 054 201-14 034 In a judo foot-sweep move, you sweep your opponent's left foot out from under him while pulling on his gi (uniform) toward that side. As a result, your opponent rotates around his right foot and onto the mat. The figure shows a simplified diagram of your opponent as you face him, with his left foot swept out. The rotational axis is through point O. The gravitational force 'g on him effectively acts at his center of mass, which is a horizontal distance d = 21 cm from point O. His mass is 69 kg, and his rotational inertia about point O is 65 kg-m2. what is the magnitude of his initial angular acceleration about point O if your pull F on his gi is (a) negligible and (b) horizontal with a magnitude of 300 N and applied at height h = 1.4 m? Assume free-fall acceleration to be equal to 9.81 m/s2 038 048 Review Score (a) Number Units Version 4.24.1.23Explanation / Answer
given horizontal distance of center of mass from the axis of rotation, d = 0.21 m
mass, m == 69 kg
moment of inertia , I = 65 kgm^2
a. For Fa = 0
initial angular acceleration alpha
I*alpha = mg*d
alpha = mg*d/I = 69*9.81*0.21/65 = 2.1868 rad/s/s
b. Fa = 300 N, h = 1.4 m
I*alpha = mg*d + Fa*h
alpha = (69*9.81*0.21 + 300*1.4)/65
alpha = 8.648 rad/s/s