A student on a piano stool rotates freely with an angular speed of 3.05 rev/s .

Question

A student on a piano stool rotates freely with an angular speed of 3.05 rev/s . The student holds a 1.45 kg mass in each outstretched arm, 0.739 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kgm2 , a value that remains constant.

Part A

As the student pulls his arms inward, his angular speed increases to 3.65 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points?

Part B

Calculate the initial kinetic energy of the system.

Part C

Calculate the final kinetic energy of the system.

Explanation / Answer

using conservation of angular momentum
3.05*(2*1.45*0.739^2+5.53)=
3.65*(2*1.45*d^2+5.53)

solve for d

d=sqrt((3.05*(2*1.45*0.739^2+5.53)/
3.65-5.53)/(2*1.45))

d=0.375 m

for KE
.5*I*^2
starting
I=2*1.45*0.739^2+5.53
7.11 kg m^2
=2.95*2*3.14 rad/s
1196 J

final
I=2*1.25*0.218^2+5.53
5.65 kg m^2
=3.64*2*3.14 rad/s
1475 J

if any problem to understand let me know

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