# Doubts about moment of inertia

1. Nov 13, 2009

### simpleton

1. The problem statement, all variables and given/known data
The polar ice caps contain about 2.3*10^19 kg of ice. This mass contributes essentially nothing to the moment of inertia of Earth because it is located at the poles, close to the axis of rotation. Because of Global warming, all the ice has melted and formed a uniform coat of water around the surface of the Earth. Calculate the change in the length of the day.

Given:
Mass of Earth = 5.98*10^24 kg
Radius of Earth = 6370 km
1 day = 86400s

2. Relevant equations
I of shell of sphere = 2/3*M*R^2
I of solid sphere = 2/5*M*R^3

3. The attempt at a solution
Actually, I have the solution to the problem, however, I don't agree with the solution, so I would like to know what you guys think.

Assuming that Earth is a sphere before and after the melting of ice, and that the mass of Earth is M, the mass of ice is m, the radius of Earth is R, then we can formulate the following equation:

Initial I = 2/5*M*R^2
Final I = 2/3*m*R^2 + 2/5*M*R^2
Initial Time = 86400s
Final time to (Final I)/(Initial I)*T -T = 0.55s

However, I don't agree with the part for the calculation of the final I. I think it should be:

2/3*m*R^2 + 2/5*(M-m)*R^2

This is because you cannot count the mass twice in my opinion. However, they said that the ice does not contribute to the moment of inertia, so I am not sure how you go around handling this :(

2. Nov 13, 2009

### tiny-tim

Hi simpleton!

(try using the X2 tag just above the Reply box )
Yes, we have to compare 2/5*(M-m)*R2 before with 2/3*m*R2 + 2/5*(M-m)*R2 after …

but the ratio m/M is about 10-5, so does it really make any difference whether we use (M-m) or just M, to 2 or 3 sig figs?

3. Nov 13, 2009

### simpleton

Yes it does.

If I use the sample method, I get 0.55s. If I use my method, I get 0.22s.

And you said that we need to compare 2/5*(M-m)*R2 before with 2/3*m*R2 + 2/5*(M-m)*R2 after …

But that does not make sense as well. The mass is different :(.

Last edited: Nov 13, 2009
4. Nov 14, 2009

### tiny-tim

(just got up :zzz: …)

But the answer is 5m/3(M - m) … how can an error ~ 10-5 make such a large difference?

(and the mass of the sphere beforehand is M-m, isn't it?)

5. Nov 14, 2009

### simpleton

Well, a day consist of 86400s, and 0.55s or 0.22s is around 10^-5 of 86400, so it does make a difference.

Anyway, if you dont believe me, here are the workings below:

(Final I)/(Initial I)*T -T

= ( 2/3*m*R^2 + 2/5*M*R^2) / ( 2/5*M*R^2 ) * T - T
= (((((2 / 3) * 2.3 * (10^19) * (6 370 000^2)) + ((2 / 5) * 5.98 * (10^24) * (6 370 000^2))) / ((2 / 5) * 5.98 * (10^24) * (6 370 000^2))) * 86 400) - 86 400
= 0.553846154s