Question

A student sitting on a frictionless rotating stool has rotational inertia 0.96 kg⋅m2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 7.35 rad/s and has negligible mass. The student extends her arms until her hands, each holding a 5.0 kg mass, are 0.80 m from the rotation axis. Ignoring her arm mass, what's her new rotational velocity? Repeat if each arm is modeled as a 0.80 m long uniform rod of mass of 4.5 kg and her total body mass is 60 kg .

Answer 1

here,

initial moment of inertia , I0 = 0.96 kg.m^2

initial angular speed , w0 = 7.35 rad/s

the final moment of inertia , I = I0 + 2 * 5 * 0.8^2

I = 0.96 + 6.4 kg.m^2 = 7.36 kg.m^2

let the new rotational speed be w

using conservation of angular momentum

I * w = I0 * w0

7.36 * w = 0.96 * 7.35

solving for w

w = 0.96 rad/s

when rod of mass m1 = 4.5 kg in introduced

the final moment of inertia , I1 = I0 + 2 * 5 * 0.8^2 + 2 * m1 * 0.8^2 /3

I = 0.96 + 6.4 kg.m^2 + 1.92 = 9.28 kg.m^2

let the new rotational speed be w1

using conservation of angular momentum

I1 * w1 = I0 * w0

9.28 * w1 = 0.96 * 7.35

solving for w1

w1 = 0.76 rad/s