Question

A person holds a 3-kg weight in each hand with arms outstretched while standing at rest on a lightweight platform that is free to turn about a vertical axis. Other people then set the person turning at a rate of one turn per 7.2 s. The person then lowers the weights until they are alongside the person's legs. At this point, the person is spinning once every 1.2 s. Estimate the person's moment of inertia.

Answer is 0.7 kg*m^2 but I don't know how to show the work!

Answer 1

initial moment of inetia of the system, I1 = I + 2*m*r^2

I = moment of inertia of the system

2*m*r^2 = moment of inertia of the masses

r = average length of the human arm = 0.76 m

initial angular speed = w1 = 2*pi/T1

after the weights are alonside of the legs

moment of inertia of masses = 0

moment of inertia of the system = I2 = I

final angular speed = w2 = 2*pi/T2

from law of conservation of angular momentum

totla angular momentum remains constant

I1*w1 = I2*w2

(I+(2*3*0.76^2))/7.2 = I/1.2

I = 0.7 kg m^2 <<---answer