Question

please halp with 5 and 6!

A person with mass mp - 70 kg stands on a spinning platform disk with a radius of R = 2.1 m and mass mg 187 kg. The disk is initially spinning at a 2 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.7 m from the center). 1) What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? 721.035 kg-m? Submit 2) What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? 446.635 kg-m2 Submit 3) What is the final angular velocity of the disk? 3.229 rad/s Submit

Answer 1

Answer: 5 centripetal acceleration of person When She is at R. a=wr wt angular velocity when she is R = 3.229 rad's - r t distance center of disk from a = (3.229) * (3:1) la=7.2985 mis? Answer

la~ 7.3 m/s² Answer Answer:6 If the persom now works I back to the rim of disk, then moment of inertia of system § 2 become If I Marisk R + mR Igo 182 7o) (2.0) If= 721.035 kg me initial moment of inertia when she was at æ , I; = 446.635 kg m? angular velocity 04, ķ, W;= 3.??9 Sadis

Using conservation of angular momentum I, W; - If Wf We = 446.635 X 3.229 7210035 We = 2.00 radis Answer so it will attain back its angular -speed,