
#1
Feb2712, 06:10 AM

P: 209

Hey,
If initially I have some solid sphere spinning at some initial angular velocity and in its final state I have the same solid sphere spinning at a different angular velocity except some of its mass has moved to a ring 45 degrees in latitude from centre , such that this ring of mass is closer to the rotational axis (rotational axis is just a diameter axis through centre of solid sphere), how do I determine the new angular velocity? I am given the radius of gyration of the solid sphere, but I'm not sure what it is and what use it would have in this scenario. I'm fairly sure initially I have some angular momentum 2/5Mr^2 * initial angular velocity, but then I'm unsure how to formulate the final state angular momentum! Thanks, S 



#2
Feb2712, 08:21 AM

PF Gold
P: 79

R = √ (I[itex]_{o}[/itex] / M) As for the initial and final angular momentum, it seems implied that momentum is conserved, but it's hard for me to tell because your description is a little bit unclear. Do you have the actual problem statement? 



#3
Feb2712, 08:29 AM

P: 209

Okay, I believe I understand what the radius of gyration is now, thanks! I'm still not sure how to apply it to the problem though.
Here is the problem statement: About 10000km3 of water is held behind dams in reservoirs around the world. Most reservoirs are at midlatitudes, 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. We are given the radius of earth, density of earth, density of water, radius of gyration. Thanks 



#4
Feb2712, 04:52 PM

P: 209

Angular Momentum  Spherical Mass & Radius of Gyration
I computed the day to be 6.03*10^5 seconds faster, is this right?




#5
Feb2712, 08:00 PM

PF Gold
P: 79

Is the radius of gyration for the earth without counting the water, or for the earth including the oceans before the redistribution to reservoirs, or for the earth including the oceans and redistributed water, or for just the oceans, etc etc. Which distribution and mass is the radius of gyration given for?
By the way, when is your problem solution due? 



#6
Feb2812, 04:53 AM

P: 209

It's due tomorrow, I was unsure as well as to what distribution the radius of gyration is given for  I've been assuming it was just for the oceans. However  this is probably vital :
The earth has mean density 5.5×103 kgm−3, radius R = 6400km and radius of gyration 0.58R. The density of water is 1gcm−3, and, for the purposes of this question, the density of sea water is not significantly different. Not sure what to infer from that in reference to the radius of gyration. All I know is that the redistribution will cause the earth to spin slightly faster, so the moments of inertia in the final state are going to be slight bigger than the initial state  it's just determining what the moment of inertia is in the final state. I'm wasn't sure if for the initial state I had to have the sum of two terms  the angular momentum of the earth given by 0.4MR^2 and the angular momentum of the oceans around the equator m(0.58R)^2 ? Though the second term will be much smaller. Thanks for helping as well! S 



#7
Feb2812, 07:38 AM

PF Gold
P: 79

Based on the additional info you gave above, the radius of gyration appears to apply to the Earth, inclusive of the oceans. If you have the slightest doubt, though, it's always best to check with your professor.
When given the radius of gyration for any body relative to some axis, the moment of inertia is always I = m[itex]\cdot[/itex]r[itex]^{2}_{gyr}[/itex] This info being given to you makes the problem easier for you. When the mass of water is redistributed to the reservoirs, you're apparently supposed to assume they're all located at midlatitudes. Here, the new moment of inertia for the redistributed water would be that of a ring of water at the same distance from the Earth's axis as the reservoirs are. That should be m[itex]_{reservoirs}[/itex][itex]\cdot[/itex][R[itex]\cdot[/itex]cos(45°)][itex]^{2}[/itex]. You would also need to recalculate the moment of inertia of the earth without the same water at it's equator. Keep in mind that the final moment of inertia will be the sum of the two moments of inertia that you calculated. So you can compare notes with another student who faced the same problem as you, check out this thread. 



#8
Feb2812, 08:06 AM

PF Gold
P: 79

By the way, the radius of gyration given to you is not the radius of gyration for a solid homogeneous sphere. This indicates that the radius of gyration given to you takes into account the actual mass distribution of the Earth; so don't calculate the moment of inertia of the Earth using equations for spheres in charts or tables.
Also, the moment of inertia of the Earth after the water has been removed from the equator is equal to I[itex]_{earth, less water}[/itex] = I[itex]_{earth}[/itex]  m[itex]_{reservoirs}[/itex][itex]\cdot[/itex]R[itex]^{2}[/itex] Make sure you understand why this is the case. 



#9
Feb2812, 08:31 AM

P: 209

Thanks a lot, would the new moment of inertia of the earth without the waters just be:
(Mass of Earth  Mass of Waters)*radius of gyration^2 or would it be (mass of earth)*radius of gyration^2  (mass of water)*R^2 so basically, would I model the waters as a ring around the equator as well? I got 2.135*10^4 s, which seems to be quite close to the other result. Thanks again! 



#10
Feb2812, 09:21 AM

P: 209

Thanks alot! I'm fairly confident I understand how to answer this question now!
Thanks again for your time, S 


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