A student sitting on a frictionless rotating stool has rotational inertia 0.97 k

Question

A student sitting on a frictionless rotating stool has rotational inertia 0.97 kg*m^2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 7.30 rad/s and has negligible mass. The student extends her arms until her hands, each holding a 5.3 kg mass, are 0.78 m from the rotation axis. Ignoring her arm mass, what's her new rotational velocity? Repeat if each arm is modeled as a 0.78 m long uniform rod of mass of 4.8 kg and her total body mass is 56 kg. what is w2 (new rotational velocity)

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 0.085 kg m^2 Initial angular momentum = 0.97*7.3 = [0.97+(2*5.3*0.78^2)]*w = final angular momentum where w is the final angular velocity of the system So w = (0.97*7.3)/[0.97+(2*5.3*0.78^2)] =.....rad/s In the repeat calculation, the new angular velocity w' will be given by w' = (0.97*7.3)/[0.97*((56-2*4.8)/56)+{(9.6*1.56^2)[Arms are considered together as rod of 9.6g mass and length 1.56 m and the moment of inertia of the student is reduced by the factor of the reduced mass/total mass] or w' =,,,,,,rad/s

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