# Rotational mechanics, angular momentum revision

## 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 :tongue:

#### Attachments

• Problem.png
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gneill
Mentor
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.

Where did the energy come from to move the dumbbells? Does it come from the work done by the man to pull the dumbells in closer? (chemical energy stored in the man's muscles I suppose) - or maybe I'm overthinking it?

gneill
Mentor
Does it come from the work done by the man to pull the dumbells in closer? (chemical energy stored in the man's muscles I suppose) - or maybe I'm overthinking it?
No, not overthinking; That's correct. So, since mechanical energy is being added by a system-internal source, the kinetic energy will not be conserved.

Right, that makes sense! Thanks a lot!