A student sitting on a frictionless rotating stool has rotational inertia 0.99 k

Question

A student sitting on a frictionless rotating stool has rotational inertia 0.99kg?m2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 6.55rad/sand has negligible mass. The student extends her arms until her hands, each holding a 4.9kg mass, are 0.72m from the rotation axis.

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

B) Repeat if each arm is modeled as a 0.72m

long uniform rod of mass of 5.0kg and her total body mass is 59kg.

Explanation / Answer

We assume that the weights in each hand were close o the axis and hence did not contribute to given moment of Inertia of her 0.99 kg m^2

Initial angular momentum = 0.99*6.55 = [0.99+(2*4.9*0.72^2)]*w = final angular momentum where w is the final angular velocity of the system

So w = 1.068rad/s

In the repeat calculation, the new angular velocity w' will be given by

0.99*6.55 = [0.99+(2*5*0.72^2)]*w.. [Arms are considered together as rod of 9g mass and length 1.6 m and the moment of inertia of the student is reduced by the factor of the reduced mass/total mass] or

w' =1.0502 rad/s

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