Question

11. + -12 points SerCP10 8.P.016. 0/4 Submissions Used My Notes + Ask Your Teacher Many of the elements in horizontal-bar exercises can be modeled by representing the gymnast by four segments consisting of arms, torso (including the head), thighs, and lower legs, as shown in Figure b. Inertial parameters for a particular gymnast are as follows. Segment Mass(kg) Arms 6.87 Torso 34.40 Thighs 13.30 Legs 7.36 Length (m) 0.530 0.601 0.374 cg (m) 0.239 0.337 0.151 1(kg • m) 0.205 1.570 0.173 0.227 0.166 Note that ro is the distance to the center of gravity measured from the joint closest to the bar when the gymnast is hanging from the bar by his hands and the masses of the arms, thighs, and legs include both appendages. I is the moment of inertia of each segment about its center of gravity. Calculate the position of the center of gravity of the gymnast shown below. Pay close attention to the definition of rc in the table. thigh arm leg 60° torso а E E

Answer 1

Given fable part to Reg. Segment reg. (mi Arms 0,239 Toarso 0.337. 0,15 vieles cum 66 torso Thighs legs 0.227 for Aruns for Torso for thighs xcg=0 Yeg= 623gr 0.239 m. sicg= 0.337m, Yey= om. reg = 0.151 cos 60°, Yega 0.151 Sinco reg = 0.0755 Yeg = 0.131 Reg= 0.227 Yega o for leg so position to the gymnast centre of gravity of the reg = = Treg. = Odo. 337 + 0.15) Cos60° +0.075520,227 0.397 + 0.0755 + 0.227. 0, 640 ml Yeg = 1 Tg = 0+ 0.151 Ein 60° to 0,131 m