A 22.8 kg turntable with a radius of 47 cm

Question

is covered with a uniform layer of dry ice
that has a mass of 4.08 kg. The angular
speed of the turntable and dry ice is initially
0.52 rad/s, but it increases as the dry ice
evaporates.
What is the angular speed of the turntable
once all the dry ice has evaporated? Answer
in units of rad/s. 1.0 m

Answer 1

You are probably suppoased to assume that the turntable is frictionless and "coasting" (unpowered) while the dry ice evaporates. This is a rather unrealistic assumption

Anyway, in that case, you should assume that angular momentum is conserved.
I1 w1 = I2 w2.

If the dry ice is evenly distributed, the moment of inertia I is proportional to total mass, M. Therefore
M1 w1 = M2 w2.
w2 = (22.8)/(22.8-4.7) x 0.52

Answer 2

You are probably supposed to assume that the turntable is frictionless and "coasting" (unpowered) while the dry ice evaporates. This is a rather unrealistic assumption

Anyway, in that case, you should assume that angular momentum is conserved.
I1 w1 = I2 w2.

If the dry ice is evenly distributed, the moment of inertia I is proportional to total mass, M. Therefore
M1 w1 = M2 w2.
w2 = [(22.8 + 4.7)/(22.8)] x 0.52

Answer 3

If you model the turntable-ice mixture, then the moment of inertia will be proportional to mass.

wf=Iinitial/I final* wi

=k(4.08+22.8)/k(22.8) * .52 rad/sec