Calculate the Final angular speed of the student?

Question

A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 0.8 m from the axis of rotation, and he rotates with angular speed of 0.72 rad/sec. The moment of inertia of the student plus the stool is 2 kgm^2 and is assumed to be constant. The student
then pulls the objects horizontally to a radius 0.38 m from the rotation axis.

** Calculate the finnal angular speed of the student. Answer in units of rad/s.

** Calculate the change in kinetic energy of the system. Answer in units of J.

Answer 1

The starting moment of inertia, I is
2+2*2*.8^2
the ending I is
2+2*2**.8^2
conservation of momentum is
(2+2*2*.8^2)*.72=
(2+2*2*.38^2)*w

w=(2+2*2*.8^2)*.72/
(2+2*2*.38^2)

=1.27 rad/s

Kinetic is .5*I*w^2
starting
.5(2+2*2*.8^2)*.72^2
1.18 J
ending
.5*(2+2*2*.38^2)*1.27^2
2.08 J

Do you know where the additional energy came from?
The work done to pull the 2 kg masses in against the centripetal force.


j

Answer 2

Consevation of angular momentum I(f)W(f) = I(i)W(i) the position; I(i) = I(o) + m(R)^2 + m(R)^2 = 3 + (a million)(.5)^2 + (a million)(.5)^2 = 3.5 I(f) = I(o) + m(r)^2 + m(r)^2 = 3 + (a million)(.35)^2 + (a million)(.35)^2 = 3.245 Then; (3.245)W(f) = (3.5)(.sixty six) W(f) = .seventy one rad/s b) KE(i) = (a million/2)I(i)(W(i))^2 = (a million/2)(3.5)(.sixty six)^2 = .seventy six j KE(f) = (a million/2)I(f)(W(f))^2 = (a million/2)(3.245)(.seventy one)^2 = .80 2 j equipment useful residences ,06 j of kinetic ability from the muscular paintings performed in pulling contained in the thousands.