A student is holding a 3.00kg weight in each hand sits on a stool that can freel
ID: 1471496 • Letter: A
Question
A student is holding a 3.00kg weight in each hand sits on a stool that can freely rotate. Assume the distance from the weights to the rotational axis of the students body is 1.00m with arms fully extended sideways at shoulder height. The lecturer gives the student a push so that the rotation of student, weight and stool is 0.750rad/s when the student's arms are fully extended. The moment of inertia of the student plus stool is 3.00kg/m2. Find the kinetic energy of the rotating system before and after the students pulls the weights inward.
Explanation / Answer
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Angular momentum when hands were extended
Inertia of student plus stool = 3.0 kg-m^2
Inertia of both weights = mr^2+ mr^2 = 2* 3.0 x 1^2 = 6.0 kg-m^2
total inertia of body = 6.0 + 3.0 = 9.0kg-m^2
therefore,
Angular momentum = Inertia x omega
Inertia = 9.0 kg-m/s
omega = 0.75 rad/s
angular momentum when hands were etended =9.0 x 0.75 = 6.75 kg-m^2/s
angular momentum after hands were pulled inwards
Inertia of student & stool = 3.0 kg-m^2
Inertia of weights = mr^2
=3.1 x 0.0^2 = 0.0 kg-m^2
total moment of inertia = 3.0 + 0.0 = 3.0 kg-m^2
Applying conservation of momentum
Angular momentum when hands were extended outwards = angular momentum when hands were pulled inside
6.75 kg-m^2/s = 3.0 x omega
angular velocity after hands were pulled inside = 6.75/3.0 =2.25 rad/s^2
now,
Rotational KE = 1/2I^2
Rotational KE when hands were extended
= 0.5 x 9.0 x 0.75^2 = 2.5312 J
Rotational KE when hands were pulled inward
= 0.5 x 3. x 2.25^2 = 7.59375 J
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