- #1
alex3
- 44
- 0
Homework Statement
Verbatim from the problem:
"About [tex]10000km^3[/tex] of water is held behind dams in reservoirs around the world.Most reservoirs are at mid-latitudes, whilst the bulk of the world’s oceans are concentrated near the equator. By using conservation of angular momentum, estimate by how much the overall movement of water into reservoirs has changed the length of the day."
Given [tex]\rho_{E}[/tex] (density of the Earth), [tex]R_{E}[/tex] (radius of the Earth) and radius of gyration [tex]0.58 R_{E}[/tex]. The density of water is given as [tex]1 g cm^{-3}[/tex], and we assuming the same for seawater.
Homework Equations
[tex]r^{2}_{g} = \frac{I}{M}[/tex]
[tex]\rho_E = \frac{M}{V}[/tex]
[tex]V = \frac{4}{3} \pi R^{3}_{E}[/tex]
[tex]L = I \omega = r^{2}_{g} M \omega[/tex]
The Attempt at a Solution
I have a feeling it's far more complicated, but I thought we have two relations for the angular momentum [tex]L[/tex]. The angular momentum must remain constant, but the moment of inertia will change depending on the distribution of mass of the Earth, so the angular frequency must change. So:
[tex]I_{s} \omega_{i} = r^{2}_{g} M \omega_{f}[/tex]
Where [tex]\omega_{i}[/tex] and [tex]\omega_{f}[/tex] represent the initial and final angular frequencies, respectively. I assume that the MoI before the water is dammed is [tex]I_{s}[/tex], and the given radius of gyration is calculated after the water is distributed. [tex]I_{s}[/tex] is the moment of inertia of a sphere,
[tex]I_{s} = \frac{2}{5} M R^2[/tex]
Then, simplifying:
[tex]\omega_{f} = \frac{2}{5 . (0.58)^{2}} \omega_{i}[/tex]
But this gives quite a large difference, the coefficient is ~1.19. I'm assuming there's a better method, but I can't fathom it.