# Inertia problem help

1. If the Earth's axis of rotation were located at a position tangent to the equator, how much more rotational intertia (moment of inerti) would the Earth have?

2. I (of sphere) = mr^2

3. I have absolutely no idea where to start.

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gneill
Mentor

Look up "Parallel Axis Theorem" in your text.

We don't use our books in class as far as the lessons, so I looked online and is this equation: I=Md^2 the same thing you are talking about?

gneill
Mentor

We don't use our books in class as far as the lessons, so I looked online and is this equation: I=Md^2 the same thing you are talking about?
That looks incomplete. It should include the original centroidal moment of inertia (about a principle axis).

If Ic is the original moment of inertia about some axis, then the moment of inertia about an axis parallel to the original axis but offset by distance d would be I = Ic + Md2, where d is the offset distance and M the mass of the object.

Okay so I think I'm starting to understand. In part 1 of the problem we found the moment of inertia using a mass of 6x10^24 and a radius of 6.378x10^3 and the formula I=mr^2 which gave me 9.76x10^31. For this part I would just take 9.76x10^31 + (6x10^24)(12756)^2 to get a final answer of 1.073893216E33?

gneill
Mentor

You'll want to be careful about your units. What units was the radius given in? Also, check your formula for the moment of inertia of a sphere -- differently shaped bodies have different expressions (at least different constants of proportionality).

Other than those things, I don't see a problem with your method.