a MasteringPhysics: Chapter '1 X + 6 C i session.masteringphysicscomi .' :l ii men iPi alvlen' l D: (Chapter 11: Rotational Dynamics and Static Equilibrium Problem 11.58 II Review V PartA A student on a piano stool rotates freely with an angular speed of 3.05 rev/s . The student holds a 1.05 kg mass in each outstretched arm, 0.779 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg . m2 , a value that remains constant As the student pulls his arms inward, his angular speed increases to 3.60 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to he points? d: 3.60 m Submit Previous Answers Muesl Answer X Incorrect; Try Again V PartB Calculate the initial kinetic energy 01 the system. bmit Lequest Answer V Partc a MasteringPhysics: Chapter 'i X + 6 C i session.masteringphysicscoml "in'ianiPit'ilwlmTlD:l (Chapter 11: Rotational Dynamics and Static Equilibrium Problem 11.58 A student on a piano stool rotates freely with an angular speed of 3.05 rev/s . The student holds a 1.05 kg mass in each outstretched arm, 0.779 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg . m2 , a value that remains constant II Review X Incorrect; Try Again V PartB Calculate the initial kinetic energy 01 the system. s n o 7 K1 = J m muest Answer V Part C Calculate the final kinetic energy of the system. ? Kf : J Muest Answer Provide Feedback Next > Page No. Date Initial K.E Ki = 1 x ( 5 . 5 3 + 2 X 1. 05 X 1 0 . 7 7 9 ) 2 ) X (3 . 05 X 270 ) 2 60 = 1 ( 5. 53 + 1. 27 43 ) x ( 0 . 10201) 2 Ki = 0 . 3 4 7 5 J Final K . E Kt = 1 x / 5 5 3# + 2 X 1 . 05 X ( 0 . 3 3 4 3 ) X 3 . 60 X 2 2 60 0 - 1 4 212 = JX (5 53+ 0. 2346 ) 2 KJ = 0 . 4009 K+ = 0 . 401 J