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

Question

A student on a piano stool rotates freely with an angular speed of 3.05 rev/s. The student holds a 1.50 kg mass in each outstretched arm. 0.749 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.43 kg middot m^2, a value that remains constant. Part A As the student polls his arms inward, his angular speed increases to 3.80 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

w1 = 3.05 rev/s , w2 = 3.8 rev/s

Ips = 5.43 kgm^2

m1 =m2 = 1.5 kg. r = 0.749 m

from conservation of angular momentum

I1w1 = I2w2

(5.43 +2*1.5*0.749^2)3.05 = (5.43 +2*1.5*d^2)3.8

d = 0.305 m

Part B:

Ki = 0.5*I1*w1^2

= 0.5*(5.43 +2*1.5*0.749^2)*(3.05*2*3.14)^2

Ki = 1305 J

Part C:

Kf = 0.5I2w2^2

= 0.5*(5.43 +2*1.5*0.305^2)*(3.8*2*3.14)^2

Kf = 1625.6 J

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