A student sitting on a frictionless rotating stool has rotational inertia .95 kg
ID: 1353203 • Letter: A
Question
A student sitting on a frictionless rotating stool has rotational
inertia .95 kg*m^2 about a vertical axis through her center of
mass when her arms are tight to her chest. The stool rotates at 6.8 rad/s
and has negligible mass. The student extends her arms
until her hands, each holding a 5.0-kg mass, are 0.75 m from the
rotation axis. (a) Ignoring her arm mass, what s her new rotational
velocity? (b) Repeat if each arm is modeled as a 0.75-m-long uniform
rod of mass of 5.0 kg and her total body mass is 65 kg.
Explanation / Answer
a) We assume that the weights in each hand were close to the axis and hence did not contribute to given moment of Inertia of her 0.95 kg m2
Initial angular momentum = 0.95 x 6.8 = [0.95+(2 x 5 x 0.752)]*w = final angular momentum
where w is the final angular velocity of the system
So w = (0.95 x 6.8)/[0.95+(2 x 5 x 0.752)] = 0.983 rad/s
b) In the repeat calculation, the new angular velocity w' will be given by
w' = (0.95 x 6.8)/[0.95 x (55/65)+(2 x 5 x 0.752+2 x 5 x 0.752/3}... [ the moment of inertia of the student body is reduced by the factor of the reduced mass/total mass] or
w' = 0.778 rad/s
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