Please show all steps. I am having a hard time solving these questions. A studen
ID: 1397673 • Letter: P
Question
Please show all steps. I am having a hard time solving these questions.
A student sitting on a chair on a circular platform of negligible mass rotates freely on an air table at initial rotational speed
2.5 rad/s . 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 kgm2 and her final rotational inertia is 5.0 kgm2 .
A.Determine the student's final rotational speed.
Express your answer to two significant figures and include the appropriate units.
B.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.
C.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.
D.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
given,
initial rotational speed = 2.5 rad/sec
weight of dumbbells = 6 kg
initial distance of dumbbells from body = 0.8 m
final distance of the dumbbells from body = 0.1 m
initial rotational inertia = 6 kg.m^2
final rotational inertia = 5 kg.m^2
By Conserve angular momentum:
initial momentum = final momentum
angular momentum = I *
initial momentum = (6 + 2 * 6 * 0.80²) * 2.5
final momentum = (5.0 + 2 * 6 * 0.10²) *
(6 + 2 * 6 * 0.80²) * 2.5 = (5.0 + 2 * 6 * 0.10²) *
a) student's final rotational speed = 6.68 rad/s
KE = ½*I*²
initial KE of the system = ½ * (6 + 2*6*0.80²) * 2.5²
initial KE of the system = 42.75 J
final KE of the system = ½ * (5.0 + 2 * 6 * 0.10²) * 6.68²
final KE of the system = 114.23 J
change in KE KE = final KE - initial KE
KE = 114.23 - 42.75
KE = 71.48 J
b) change of kinetic energy of the system = 71.48 J
initial KE = ½ * (2*6*0.80²) * 2.5²
initial KE = 24 J
final KE = ½ * (2*6*0.10²) * 6.68²
final KE = 2.68 J
KE = 2.68 - 24
KE = -21.32 J
c) the kinetic energy of the system = -21.32 J
initial KE = ½ * 6 * (2.5)²
initial KE = 18.75 J
final KE = ½ * 5 * (6.68)²
final KE = 111.556 J
KE = 111.556 - 18.75 J
KE = 92.806 J
d) change of kinetic energy of the system = 92.806 J
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