(1/2) of the problem is done: Need to find part (c) and (d) A 6.4kg turntable is

Question

(1/2) of the problem is done: Need to find part (c) and (d)

A 6.4kg turntable is free to rotate on its axis. The turntable has a radius of 0.60m and is rotating at a rate of 52rad/s.

I FOUND: The rotational inertia of the turntable to be 1.15kgm^2 (correct)
I FOUND: The angular momentum of the turntable to be 59.8kgm^2/s (correct)

I cannot figure out part (c) and (d)

(c) A 2.50kg piece of clay falls directly down onto the turntable at a position of 0.35m from the center and sticks to the surface. What is the final angular velocity?

(d) What is the change in kinetic energy?


Need help finishing! Thank you!

Explanation / Answer

angular momentum must be conserved.

initial momentum is 59.8kg*m^2/s
the end result must be
(moment of inertia for the turntable + the moi for the clay)*w
moi of the turntable = 1.15 kg*m^2
moi clay = mr^2(treat it as a point mass) = 2.5 kg *.35m^2 = .306 kg*m^2/s

now L = (moi turntable + moi clay) * w
59.8 = (1.15 + .306)w
w = 41.07 rad/s

b.
Ke = 1/2 Iw^2
initial Ke = .5(1.15)*(52^2) = 1.55kJ
final ke = .5(1.15 + .306)(41.07)^2 = 1.23kJ

change in ke = kef - kei = 1.23kJ - 1.55kJ = -320J

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