A student on a piano stool is rotating freely with an angularspeed of 3.05 rev/s
ID: 1724746 • Letter: A
Question
A student on a piano stool is rotating freely with an angularspeed of 3.05 rev/s. the student holds a 1.50 kg mass in eachoutstretched arm, 0.759m from the axis of rotation. The combinedmoment of inertia of the student and the stool, ignoring the twomasses is 5.53 kg m^2, a value that remains constant. A) As the student pulls his arms inward, his angular speedincreases to 3.69 rev/s. How far are the masses from the axis ofrotation? B)What is the final and inital KE of the system? A student on a piano stool is rotating freely with an angularspeed of 3.05 rev/s. the student holds a 1.50 kg mass in eachoutstretched arm, 0.759m from the axis of rotation. The combinedmoment of inertia of the student and the stool, ignoring the twomasses is 5.53 kg m^2, a value that remains constant. A) As the student pulls his arms inward, his angular speedincreases to 3.69 rev/s. How far are the masses from the axis ofrotation? B)What is the final and inital KE of the system?Explanation / Answer
I1 = 5.53 + 2(1.5)(0.759)^2= 7.2582kgm^2
According to conservation of angular momentum
I22 =I11
I2= I11 / 2
I2 = 7.2582(3.05) /3.69
= 5.9993
So moment of inertia is
5.53 + 2(1.5)R^2 =5.9993
R = 0.3955m
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initial KE
Ki =(1/2)I112
= 33.759J
Final kinetic energy
Kf =(1/2)I222
= 40.8435J
= 40.8435J
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