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A turntable with a moment of inertia of 0.014 kg*m2 rotates feely at 2.7 rad/s. A circular disk of mass 500 g and diameter 22 c...Asked by Unknown
A turntable with a moment of inertia of 0.029 kg*m^2 rotates feely at 2.8 rad/s. A circular disk of mass 270 g and diameter 20 cm, and initially not rotating, slips down a spindle and lands on the turntable.
a) Find the new angular speed.
b) What is the change in kinetic energy?
a) Find the new angular speed.
b) What is the change in kinetic energy?
Answers
Answered by
drwls
a) Angular momentum about the spindle is conserved.
Iturntable*w1 = (Iturntable+Icd)w2
Solve for the final angular velocity w2.
b) (1/2)*Iturntable*w1^2 - (1/2)(Iturntable + Icd)*w2^2
The formula for the compact disc moment of inertia Icd is:
Icd = (1/2)Mcd*Rcd^2
Iturntable*w1 = (Iturntable+Icd)w2
Solve for the final angular velocity w2.
b) (1/2)*Iturntable*w1^2 - (1/2)(Iturntable + Icd)*w2^2
The formula for the compact disc moment of inertia Icd is:
Icd = (1/2)Mcd*Rcd^2
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