A student on a piano stool rotates freely with an angular speed of 2.85 rev/s .

Question

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.25 kg mass in each outstretched arm, 0.789 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.13 kgm2 , a value that remains constant.

a) As the student pulls his arms inward, his angular speed increases to 3.52 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points?

d= ____________m

b) Calculate the initial kinetic energy of the system.

Ki=___________J

c)Calculate the final kinetic energy of the system.

Kf=____________J

Explanation / Answer

(a)

using conservation of angular momentum
2.85*(2*1.25*0.789 2 + 5.13) = 3.52*(2*1.25*d2+5.13)
2.85*(1.5563 + 5.13) = 3.52*(2*1.25*d2+5.13)
19.0559 = 3.52*(2.5*d2+5.13)
5.4136 - 5.13 = 2.5*d2
d = 0.3368 m

(b)
Initial KE = ½*I*2

I = 2*1.25*0.7892 + 5.13 = 6.6863 kg m2

= 2.85*2*3.1415 = 17.9070 rad/s

KE = ½*I*2 = 1072.02 J

(c)

Final KE = ½*I*2

I = 2*1.25*0.33682 + 5.13 = 5.4135 kg m2

= 3.52*2*3.1415 = 22.1168 rad/s

KE = ½*I*2 = 1324.01 J

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