Doubts about moment of inertia

[simpleton]

Homework Statement

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

Homework Equations

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

Given Answer:
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 :(

 

[tiny-tim]

Hi simpleton!

(try using the X² tag just above the Reply box )

The polar ice caps contain about 2.3*10^19 kg of ice.
…
Mass of Earth = 5.98*10^24 kg
…
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 :(

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?

 

[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 :(.

 

[tiny-tim]

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 :(.

(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?)

 

[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 don't believe me, here are the workings below:

Given Answer:

(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

My Answer:

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

= ( 2/3*m*R^2 + 2/5*(M-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)) - (2.3 * (10^19))) * (6 370 000^2))) / ((2 / 5) * 5.98 * (10^24) * (6 370 000^2))) * 86 400) - 86 400
= 0.221538462

And tiny-tim, your answer is actually the same as the sample answer given if you multiply your answer (6.41028107 × 10^-6) by 86400. If you express your answer in days, it does not make a difference. But if it is in SI units, there is a difference of around 0.30 seconds.

Can you tell me why my method is wrong?