Fill This Form To Receive Instant Help
Fill This Form To Receive Instant Help
Homework answers / question archive / A student on a piano stool rotates freely with an angular speed of 3
A student on a piano stool rotates freely with an angular speed of 3.16 rev/s. The student holds a 1.50 kg mass in each outstretched arm, 0.794 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.45 kg*m2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.46 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?
Calculate the initial kinetic energy of the system.
Calculate the final kinetic energy of the system.
Be sure to show all your steps and calculations. This problem has three parts so you should have three different answers for each part.
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.
