A student on a piano stool rotates freely with an angular speed of 3

Please see the attached file.

Solution

As the student is not subjected to any external torque, his angular momentum remains same with his hands outstretched and his hands pulled in.

Moment of inertia of the student + stool + masses with hands outstretched = I0 = 5.45 + 2 x 1.5 x (0.794)2 = 7.34 kgm2

Angular velocity with hands outstretched = ω0 = 2Πf0 = 2Π x 3.16 = 19.85 rad/sec

Angular momentum with hands outstretched = I0ω0 = 7.34 x 19.85 = 145.7 kgm2/sec

Angular velocity with hands pulled in = ω1 = 2Πf1 = 2Π x 3.46 = 21.74 rad/sec

Let the masses be at a distance l from the axis of rotation with hands pulled in. Then, applying the principle of conservation of angular momentum we can write :

I1ω1 = I0ω0

[5.45 + 2 x 1.5 x l2] x 21.74 = 145.7

Solving we get : l = 0.646 m

Initial kinetic energy = ½ I0(ω0)2 = ½ x 7.34 x 19.852 = 1446 J

Final kinetic energy = ½ I1(ω1)2

I1 = [5.45 + 2 x 1.5 x 0.6462] = 6.70 kgm2

Final KE = ½ x 6.7 x 21.742 = 1583.3 J.