- #1
Nono713
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MHB
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Homework Statement
The problem is attached as an image. Note this is from a past exam.
Homework Equations
Conservation of angular momentum.
Rotational kinetic energy.
The Attempt at a Solution
a) The moment of inertia of the man and stool is given as 3 kg m^2, and the dumbells can be considered point masses, so we can just add them all up:
I_\mathrm{initial} = I_\mathrm{man} + 2I_\mathrm{dumbell~~~ away} = 3 + 2(3 \times 1^2) = 9
I_\mathrm{final} = I_\mathrm{man} + 2I_\mathrm{dumbell ~~~pulled~~~ in} = 3 + 2(3 \times 0.25^2) = 3.375
b) The initial angular velocity of the man is 1.5, and its initial moment of inertia is 9, so the system's angular momentum is L = 1.5 \times 9 = 13.5. From conservation of angular momentum, the system must have the same angular momentum after the dumbells have been pulled in, so L = I_\mathrm{final} \omega_\mathrm{final}. So \omega_\mathrm{final} = \frac{L}{I_\mathrm{final}} = \frac{13.5}{3.375} = 4.
c) Using the rotational kinetic energy formula:
K_\mathrm{initial} = \frac{1}{2} I_\mathrm{initial} \omega^2_\mathrm{initial} = \frac{1}{2} \times 9 \times 1.5^2 = 10.125 J
K_\mathrm{final} = \frac{1}{2} I_\mathrm{final} \omega^2_\mathrm{final} = \frac{1}{2} \times 3.375 \times 4^2 = 27 J
I am note sure I got it right, shouldn't kinetic energy be conserved? Or does some of it go into potential energy because of the increased radius? I mean clearly the guy is going to spin faster, so the extra kinetic energy must be coming from somewhere.
PS: imagine the correct units are in there, I'm just too lazy to type them up in LaTeX
