A girl is rotating on a piano stool with an angular speed of 2.95 rev/s (assume

Question

A girl is rotating on a piano stool with an angular speed of 2.95 rev/s (assume no friction). She holds a 1.25-kg mass in each of her outstretched arms, 0.759 m from the center of her body (i.e. the axis of rotation). The combined moment of inertia of the girl's body (excluding her arms and the two masses) and the stool, is 5.43 kg middot m^2, and remains constant. (a) As the girl pulls her arms inward, her angular speed increases to 3.54 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points? (b) Calculate the initial and final kinetic energy of the system.

Explanation / Answer

here,

initial angular velocity, w1 = 2.95 rev/s = 0.3089 rad/s

final angular apeed, w2 = 3.54 rev/s = 0.37 rad/s

initial moment of inertia, I1 = 5.43 + 2 * 1.25 * 0.759^2
initial moment of inertia, I1 = 6.87 kg.m^2

final moment of inertia, I2 = 5.43 + 2 * 1.25 * x^2

From conservation of momentum :
before = after

I1 * w1 = I2 * w2

2.95 * 6.87 = 3.54 * (5.43 + 2 * 1.25 * x^2)

distance of masses, x = 0.344 m from axis of rotation

initial Kinetic energy, kei = 0.5*I1*w1^2
initial Kinetic energy, kei = 0.5*6.87*0.3089^2
initial Kinetic energy, kei = 0.328 J

Final Kinetic energy, kef = 0.5*I2*w2^2
Final Kinetic energy, kef = 0.5*(5.43 + 2 * 1.25 * 0.344^2)*0.37^2^2
Final Kinetic energy, kef = 0.054 J

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