A student on a piano stool rotates freely with an angular speed?

A student on a piano stool rotates freely with an angular speed of 2.95 rev/s. The student holds a 1.25 kg mass in each outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg *m^2, a value that remains constant.


As the student pulls his arms inward, his angular speed increases to 3.64 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?


find d, KE initial and KE final

A student on a piano stool rotates freely with an angular speed?
using conservation of angular momentum


2.95*(2*1.25*0.759^2+5.53)=


3.64*(2*1.25*d^2+5.53)





solve for d





d=sqrt((2.95*(2*1.25*0.759^2+5.53)/


3.64-5.53)/(2*1.25))





d=0.218 m





for KE


.5*I*ω^2


starting


I=2*1.25*0.759^2+5.53


6.97 kg m^2


ω=2.95*2*3.14 rad/s


1196 J





final


I=2*1.25*0.218^2+5.53


5.65 kg m^2


ω=3.64*2*3.14 rad/s


1475 J





The additional energy came from the work the student did to pull in the arms and weights.