A student sits on a freely rotating stool holding two weights, each of mass 5 kg

Question

A student sits on a freely rotating stool holding two weights, each of mass 5 kg. When his arms are extended horizontally, the weights are 1 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3 kg*m^2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.3 m from the axis of rotation 1). Find the new angular speed of the student? 2). Find the kinetic energy of the rotating system (comprised of student, stool, and weights) before and after he pulls the weights inward?

Explanation / Answer

(a) Find the new angular speed of the student. in rad/s Angular momentum before = angular momentum after Angular momentum before Inertia of student plus stool = 3.1 kg-m^2 Inertia of weights = mr^2 = 3.1 x 1^2 = 3.1 kg-m/s 3.1 x 2 = 6.2 6.2 + 3.1 = 9.3 kg-m^2 Angular momentum = Inertia x omega Inertia = 9.3 omega = 0.754 rad/s 9.3 x 0.754 = 7.0122 kg-m^2/s angular momentum after Inertia of student & stool = 3.1 kg-m^2 Inertia of weights = mr^2 3.1 x 0.29^2 = 0.26071 kg-m^2 0.26071 x 2 = 0.52142 3.1 + 0.52142 = 3.62142 7.0122 kg-m^2/s = 3.62142 x omega 7.0122/3.62142 = omega = 1.9363 rad/s (answer) (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward. in J (before) in J (after) Rotational KE = 1/2I?^2 Before = 0.5 x 9.3 x 0.754^2 = 2.6436 J After = 0.5 x 3.62142 x 1.9363^2 = 6.7888 J

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