A student sits on a freely rotating stool holding two dumbbells, each of mass 2.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 2.99 kg (see figure below). When his arms are extended horizontally, the dumbbells are 1.07 m from the axis of rotation and the student rotates with an angular speed of 0.740 rad/s. The moment of inertia of the student plus stool is 2.60 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.291 m from the rotation axis. (a) Find the new angular speed of the student. rad/s (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.

Explanation / Answer

a)

Angular momentum before = angular momentum after

Angular momentum before

Inertia of student plus stool = 2.60 kg-m^2
Inertia of weights = mr^2 = 2.99 x 1.07^2 = 3.42 kg-m/s
3.42 x 2 = 6.84

6.84 + 2.60 = 9.44 kg-m^2

Angular momentum = Inertia x omega

Inertia = 9.44
omega = 0.740 rad/s

9.44 x 0.740 = 6.99 kg-m^2/s

angular momentum after

Inertia of student & stool =2.60 kg-m^2
Inertia of weights = mr^2

2.99 x 0.291^2 = 0.2532 kg-m^2

0.2532 x 2 = 0.5064

2.60 + 0.5064 = 3.1064

6.99 kg-m^2/s = 3.1064 x omega

omega = 2.25 rad/s (answer)

(b)

Rotational KE = 1/2I^2

Before = 0.5 x 9.44 x 0.740^2 = 2.585 J

After = 0.5 x 3.1064 x 2.25^2 = 7.863 J

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