Question

A student sits on a rotating stool holding two2.7-kg objects. When his arms are extended horizontally, the objects are 1.0 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 3.0 kg

Answer 1

first find the moment of inertia before and after:

I = stool + 2 * mass * distance^2

initial: I = 3.0 + 2*2.7*1.0^2 = 8.4

final: I = 3.0 + 2*2.7*0.28^2 = 3.4234

Now...

a) final angular speed = init ang speed * init I / final I =

= 0.75 * 8.4 / 3.4234 =

= 1.84 rad/s

b) KE before = (1/2) I ^2 = (1/2)*8.4*0.75^2 =

= 2.36 J

KE after = (1/2) * 3.4234 * 1.84^2 =

= 5.80 J

Answer 2

Initial MI = 3 + 2*3.6*1^2 = 10.2 kg m^2 .. Final MI = 3 + 2*3.6*.48^2 = 4.66 kgm^2 ....(a) Applying conservation of angular momentum, 10.2*.75 = 4.66*Wf on solving, we get final angular speed = Wf = 1.64 rad/s .....(b) Initial K.E = .5*MI*Wi^2 = .5*10.2*.75^2 = 2.87 J ...Final K.E = .5*MI*Wf^2 = .5*4.66*1.64^2 = 6.27 J

Answer 3

MI at start = 3 + 2 . 2.6 . 1^2 = 8.2 kg m^2
Angular momentum ( L ) = 8.2 . 0.75 = 6.15 kg m^2 / s

MI at end = 3 + 2 . 2.6 . 0.37^2 = 3.712 kgm^2

New angular velocity = 6.15 / 3.712 = 1.66 rad/s

Original KE = 1/2 . MI . w^2 = 0.5 . 8.2 . 0.75^2 = 2.31 J

Final KE = 0.5 . 3.712 . 1.66^2 = 5.1 J

The extra energy comes from the work the student does in pulling the masses inwards.