Question

A student sits on a rotating stool holding two 2.2-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 · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.46 m from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the student before and after the objects are pulled in. before after (c) Where does the energy difference come from/go to?

Answer 1

Data:
ω1:=0.75 rad/second; I:=3.0 kg-m²; m:=2.2 kg; d1:=1 m; d2:=0.46 m;

Use a conservation of momentum approach.

Initially, the rotational inertia of the student + stool + weights system is:
I_net1 = I + 2*m*d1²
=7.4
I is the rotational inertia of empty handed student + stool
m is the mass of one of the barbels in his hand, assumed as point masses. The leading factor of 2 shows up because there is one in each hand.
d1 is the initial positioning of each barbell

This system is rotating at angular velocity ω1.

Its angular momentum is thus:
L = I_net1*ω1

When the student pulls in the barbells, the new rotational inertia is:
I_net2 = I + 2*m*d2²
=3.93
Angular momentum after pulling in barbells:
L = I_net2*ω2

Because no external torques act on the freely rotating system, ANGULAR MOMENTUM is conserved.

Thus:
I_net1*ω1 = I_net2*ω2

Solve for ω2:
ω2 = ω1*I_net1/I_net2
=0.75*7.4/3.93

=1.4122
And compute the rotational kinetic energies:
KE1 = (1/2)*I_net1*ω1²

=2.08125
KE2 = (1/2)*I_net2*ω2²
=3.9188

So, we used conservation of angular momentum to solve this problem...
But why is kinetic energy not conserved? Where did this KE come from?

Well, the student needed to do(3.9188-2.08125) J of work to bring the change in angular velocity.