Question

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 F→g on him effectively acts at his center of mass, which is a horizontal distance d = 29 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→a 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.

Answer 1

@ Apply Newton's and law, mg = 69 (9.81). T = IL mgd=It a = 6964-8)(0:29) - 3:01]rad/82 65 (Thet) = 69x9.81X0:29 + 14 (300) - 6510) x= [9.48 rad/g2