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

Question

A student on a piano stool rotates freely with an angular speed of 2.85rev/s . The student holds a 1.30kg mass in each outstretched arm, 0.789m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.23kg*m^2 , a value that remains constant.

Part A: As the student pulls his arms inward, his angular speed increases to 3.46rev/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

initial angular momentum

L1 = (I + 2*m*r^2)*w

L1 = (5.23+(2*1.3*1.3*0.789*0.789))*(2.85) = 20.1 kg m^2/s

partA)

L2 = (I + 2*m*r1^2)*w1

L2 = (5.23+(2*1.3*1.3*r1*r1))*(3.46)

L2 = (5.23 + 3.38*r1^2)*3.46

L2 = 18.0958 + 11.6948 *r1^2

law of conservation of angular momentum

L2 = L1

18.0958 + 11.6948 *r1^2 = 20.1

r1 = 0.171 m

part B )

w = 2.85*2*3.14 = 17.898 rad/s

KEi = 0.5*I*w^2 = (5.23+(2*1.3*1.3*0.789*0.789))*(17.898*17.898)

KEi = 2349.4 J

part C

w1 = 3.46*2*3.14 = 21.73 rad/s

KEf = 0.5*I*w1^2

KEf = 0.5*(5.23+(2*1.3*1.3*0.171*0.171))*21.73*21.73

KEf = 1258.12 J

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