A student is sitting on a chair and spinning at a rate of 2.15 revolutions per s

Question

A student is sitting on a chair and spinning at a rate of 2.15 revolutions per second. The student is holding a 1.50 kg mass in each hand, stretched out at arm's length, which is a distance of 0.710 m from the center of rotation. The student then brings both arms in, speeds up, and spins at a rate of 2.50 revolutions per second. The inertia of the student and chair is 5.25 kgm2. (Treat the weights as point masses.)

a. At the higher rate of rotation, what is the distance of each mass in the student's hand from the center of rotation?

b. What is the change in kinetic energy of the student?

Explanation / Answer

a) Initially moment of inertia Ii = (5.25 kg m^2) + ( 2 x 1.50 x 0.710^2)

Ii = 6.76 kg m^2

Finally suppose masses are at a distance of d.

then final moment of inertia If = ( 5.25) + ( 2 x1.50d^2 ) = 5.25 + 3d^2

using angular momentum conservation,

Ii wi = If wf

6.76 x 2.15 = (5.25 + 3d^2 )x 2.50

3d^2 = 0.566

d = 0.434 m

b) initial KE = Ii wi^2 /2

KEi = (6.76 x (2.15 x 2pi rad/s)^2 /2) = 616.81 J

fianl KE :

KEf = (5.25 + 3(0.434^3)) ( 2.50 x 2pi)^2 /2 = 717.40 J

change in KE = KEf - KEi = 100.60 J

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