A student sitting on a frictionless rotating stool has rotational inertia 0.94 k

Question

A student sitting on a frictionless rotating stool has rotational inertia 0.94 kgm2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 7.05 rad/s and has negligible mass. The student extends her arms until her hands, each holding a 4.6 kg mass, are 0.75 m from the rotation axis.

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

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 63 kg .

Explanation / Answer

here,

initial rotational inertia , Ii = 0.94 kg.m^2

initial angular speed , wi = 7.05 rad/s

the final moment of inertia , If = Ii + 2*4.6*0.75^2

If = 6.115 kg.m^2

let the new angular velocity be wf

using conservation of angular momentum

If*wf = Ii*wi

6.115 * wf = 0.94 * 7.05

wf = 1.08 rad/s

the new angular velocity is 1.08 rad/s

when the arms are also modeled

the final moment of inertia , If' = Ii + 2*4.6*0.75^2 + (2/3) *mrod * Length^2

If' = Ii + 2*4.6*0.75^2 + (2/3) * 4.7 * 0.75^2

If = 7.88 kg.m^2

let the new angular velocity be wf

using conservation of angular momentum

If*wf = Ii*wi

7.88 * wf = 0.94 * 7.05

wf = 0.84 rad/s

the new angular velocity is 0.84 rad/s

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