Posted by Christina on Monday, April 14, 2008 at 6:00pm.
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
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
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
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