A student tiding on a chair on a circular platform of negligible matt routes fre
ID: 3162175 • Letter: A
Question
A student tiding on a chair on a circular platform of negligible matt routes freely on an air table at initial rotational speed 2.5 rad/a The student's arms are initially extended with 6.0-kg dumbbells in each hand. As the student pulls her arms in toward her body, the dumbbells move from a distance of 0.80 m to 0.10 m from the axis of rotation The initial rotational inertia of the student's body (not including the dumbbells) with arms extended is 6 0 kg middot m^2 and her final rotational menu a 50 kg middot m: Determine the student's final rotational speed. Express your answer to two significant figures and include the appropriate units. Determine the change of kinetic energy of the system consisting of the student together with the two dumbbells. Express your answer to two significant figures and include the appropriate units. Determine the change in the kinetic energy of the system consisting of the two dumbbells alone without the student Express your answer to two significant figures and include the appropriate units. Determine the change of kinetic energy of the system consisting of student alone without the dumbbells Express your answer to two significant figures and Include the appropriate units.Explanation / Answer
A. Conserve angular momentum:
(6 + 2*6*0.80²)kg·m² * 2.5rad/s = (5.0 + 2*6*0.10²) *
= 6.68 rad/s
B. initial KE = ½*I*² = ½ * (6 + 2*6*0.80²)kg·m² * (2.5rad/s)² = 42 J
final KE = ½ * (5.0 + 2*6*0.10²) * (6.68 rad/s)² = 114.2 J
KE = 72.2 J
C. initial KE = ½ * (2*6*0.80²)kg·m² * (2.5rad/s)² = 24 J
final KE = ½ * (2*6*0.10²)kg·m² * (5.3rad/s)² = 1.68 J
KE = 22.32 J
D. initial KE = ½ * 6kg·m² * (2.5rad/s)² = 18.75 J
final KE = ½ * 5kg·m² * (5.3rad/s)² = 70 J
KE = 51.25 J
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