A student sitting on a frictionless rotating stool has rotational inertia 0.96 k

Question

A student sitting on a frictionless rotating stool has rotational inertia 0.96 kgm2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 6.60 rad/s and has negligible mass. The student extends her arms until her hands, each holding a 4.9 kg mass, are 0.75 m from the rotation axis.

Part A: Ignoring her arm mass, what's her new rotational velocity?

Part B: Repeat if each arm is modeled as a 0.75 m long uniform rod of mass of 4.7 kg and her total body mass is 57 kg.

Explanation / Answer

initial moment of inertia = I1 = 0.96 kg m^2

initial angular velocity = w1 = 6.6 rad/s

final moment of inertia = I2 = I1 + 2*m*r^2 = 0.96 + (2*4.9*0.75^2) = 6.47 kg m^2

final angular velocity = w2

from conservation of angular momentum Li = Lf

I1*w1 = I2*w2

0.96*6.6 = 6.47*w2

w2 = 0.98 rad/s      <<<<<---------------------answer

++++++++++++++

part B

final moment of inertial = I1 + 2*m*r^2 + 2*(1/3)*m*r^2

I2 = 0.96 + (2*4.7*0.75^2))+((2/3)*4.7*0.75^2 = 8.01 kgm^2

I1*w1 = I2*w2

0.96*6.6 = 8.01*w2

w2 = 0.791 rad/s <<-------answer

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