A student, sitting on a stool rotating at a rate of 40 RPM, holds masses in each hand....?

Question

A student, sitting on a stool rotating at a rate of 40 RPM, holds masses in each hand. When his arms are extended, the total rotational inertia of the system is 6.5 kg·m2. He pulls his arms in close to his body, reducing the total rotational inertia to 3.5 kg·m2. If there are no external torques, what is the new rotational velocity of the system?

______RPM

Answer 1

this is a problem in conservation of angular momentum

the angular momentum is constant, so we know that

Iω (start)=Iω(end)
6.5*40=3.5*ω(end)
ω(end)=6.5*40/3.5=74.2 rpm

Answer 2

Conservation of angular momentum.

I1*w1=I2*w2
since RPM translates to rad/s by a constant, I will work in RPM

6.5*40=3.5*w2

74.3 RPM

BTW: The conservation of energy equation is complex for this since the student must do work to pull in his arms. The angular kinetic energy is increased because of the work done.

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Answer 3

This student needs to find a hobby or something.

Answer 4

by capacity of the conservation of angular momentum Iw (preliminary) = Iw (very final) the place I is the inertia and w is the angular velocity you're given the frequency f (preliminary) = 38 rpm w = 2 * pi * f = 76pi rad/min ...(preliminary) so now you have 6.5 * 76pi = 3.5 * w ....(very final) so the strategies-blowing w = 141.14(pi) rad/min w = 443.4 rad/min edit: sorry, i replaced them to rad/min like they might desire to be.