A woman sits on a stool that rotates freely with an angular speed of 3.0 rev/s.

Question

A woman sits on a stool that rotates freely with an angular speed of 3.0 rev/s. The woman holds 2.5 kg mass in each hand of her outstretched arms. Her arms about 0.80 meters from her center of rotation. The combined moment of inertia of the woman is 5.6 kg*m^2 (this does not include the moment of inertia of the weights). This value remains constant.

1. As the woman pulls her arms inward, her angular speed increases to 3.6 rev/s. How far are the masses from the axis of rotation?

2. Calculate the initial and final kinetic energies of the system.

Explanation / Answer

given

w1 = 3 rev/s = 3*2*pi rad/s = 18.85 rad/s

w2 = 3.6 rev/s = 3.6*2*pi rad/s = 22.6 rad/s

r1 = 0.8 m

r2 = ?

1) Apply conservation of angular momentum

I2*w2 = I1*w1

(5.6 + 2*2.5*r2^2)*3.6 = (5.6 + 2*2.5*0.8^2)*3

(5.6 + 5*r2^2)*3.6 = 26.4

5*r2^2 = 26.4/3.6 - 5.6

r2 = sqrt(1.733/5)

= 0.59 m

2)

KEi = (1/2)*I1*w1^2

= (1/2)*(5.6 + 2*2.5*0.8^2)*18.85^2

= 1563 J

KEf = (1/2)*I2*w2^2

= (1/2)*(5.6 + 2*2.5*0.59^2)*22.6^2

= 1874 J

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