Question

7. If the polar ice caps melt, adding more liquid water to the oceans, the Earth’s rotational inertia would increase by as much as 0.3%. Compute the effect such a change would have on the length of 1 day, by assuming the Earth’s angular momentum remains constant.

Answer 1

Well, you want to know how fast the earth would rotate compared to now, with the moment of inertia changing and the angular momentum staying the same.
Do you know how, "speed of rotation" or frequency relates to those other 2 quantities? If not, what do you think? Just take a guess - it's really not complicated..

Answer 2

I know that angular momentum for one full day for the earth is 7.07 x 10^33 kgm^2/s
and the moment of inertia equation is (2/5) MR^2. But I don't understand how speed of rotation relates to that at all

Answer 3

Well, we don't want to compute an angular momentum or the moment of inertia in this case.. We want to know, how these two relate to each other.
Hint: the term commonly used for "speed of rotation" is angular velocity (ω)

Now how do you think, do angular velocity, moment of inertia and angular momentum relate?

Answer 4

Just guess.. and remember: moment of inertia tells you, how hard it is, to change the state of motion of an object..

Answer 5

I honestly can't even tell you how they relate. I am so confused on this chapter. My notes are doing me no help at all

Answer 6

I really don't want to annoy you, but if you get there on your own, the "aha"-moment will be worth the trouble ;)

So ahm.. do you know, how regular, ordinary momentum is defined?

Answer 7

isnt that just mass x velocity?

Answer 8

That's exactly right!\[p = m \, v\]Now comes the fun part... remember, when I told you to keep in mind, that moment of inertia tells you how hard it is, to change an object's state of motion? Well.. can you think of another physical quantity that does exactly that? (hint: look at the above equation ;) )
The other 2 quantities we are trying to relate are called angular _velocity_ and angular _momentum_ ... guess what they relate to ^_^

And now try to figure out the formula again ;)

Answer 9

So the moment of inertia is I=angular velocity x angular momentum? sorry im really horrible at physics

Answer 10

What do you think correspondes to moment of inertia being the quantity, that tells you how hard it is to move something (the bigger, the quantity, the harder it gets) in you regular, non-rotational equation\[p = m \, v\]???

Answer 11

angular momentum would = I x angular velocity? If thats not it ill have to give up because ill never get it

Answer 12

Are you just guessing or really thinking about it? ;)

Just look at my last question and think about your every-day live? What do you call objects, that are hard to move? fast? heavy? high-momentuary? :P

Answer 13

force? i really dont know. im having to guess because i have no honest idea.

Answer 14

whaaaaat? xD

Well, I don't know about you, but I like to refer to objects that are hard to move, as "heavy".. as in >>God damnit, that's one heavy washing-machine<<

So the point I wanted to make was, the moment of inertia is essentially the same thing a mass - just for "rotational-stuff".. okay?

Answer 15

Angular momentum remains conserved like linear momentum

Use these relations,
\[ \huge J_1 = I_1\omega_1 = J_2 = I_2\omega_2 \]

\[ \huge \omega = 2\pi/T\]

T is time period of rotation.

Answer 16

@experimentX - please keep from just posting the answer.. it really does not help and in fact violates the code of conduct ( http://openstudy.com/code-of-conduct).

Anywho: back to our problem:
As you already guessed correctly\[L = I \, \omega\]\[p = m \, v\] See, how angular momentum correspondes to "normal"/linear momentum, moment of inertia to mass and angular velocity to "normal"/linear velocity?

Answer 17

yes i think i understand what you are saying

Answer 18

Tell me the differences and similarities between angular and linear velocities.

Answer 19

Ok, now we want to know the difference in the length of a day and as experimentX already told us\[\omega = 2\pi \, f = \frac{2 \pi}{T}\]
We want to solve for T, don't we?

Can you do it, @jknighton1 ?

Answer 20

yes ... from there you can calculate \(\omega1\) ,I1 and I2 will cancel out... find w2 using the above relation. then find T

Answer 21

I don't get what the f has to do, orwhat to put in there or anything. im so confused..

Answer 22

The f is called "frequency" - it says how often (measured in "per second") something happens. In our case, one rotation of the earth. We don't need it in this example - I just wrote this intermediate step down, so you'd see how we get to our \[\omega = \frac{2 \pi}{T}\]T is called the "period" and ω is the "angular frequency".

Answer 23

angular frequncy? i dont get where that comes from. i thought we were talking about angular momentum and angular velocity

Answer 24

oh snap.. sry :-/
That's my fault - it never was angular velocity, but angular speed

angular frequency ( or radial/circular/orbital/radian frequency) is the same as angular speed!

Angular velocity is something different - forget about it. I'm very sorry about that - I'm no native speaker and in german there is no distinction between the terms speed and velocity, so I just translated without thinking and chose the wrong term ...

Answer 25

I'm so lost. I've spent over an hour on this problem and still have 8 more to go. Im gonna give up on this one. its just too confusing

Answer 26

you need just two relations ... this should be easy.

Answer 27

so there is 86400 seconds in one day. where am i putting this 0.3 into this equation to even figure anything out. I really do not know anything about this at all its very hard for me

Answer 28

NO - don't put numbers in there... just calculate it using the ahm.. "characters"? You kow, the names of the quantities.. It is really, really simple..

Answer 29

\[\omega = \frac{2 \pi}{T}\]\[L = I \omega\]
Just plug the first equation into the second, solve for T and think about what will happen to T, if you increase I by 0.3%
You can do it! :)

Answer 30

T=(I * 2π)/L ? right. So if you increase I then it is going to make the day longer. But shouldn't I need to know how much it is going to affect on the time

Answer 31

That's correct. And yes, you should.. but remember what the problem stated about the angular momentum?

Answer 32

It stays constant. I already know the angular momentum. it is 7.08 x 10^33.

Answer 33

That's right.. So \[\frac{2 \pi}{L} = constant\]So you got some quantity T = const. * I ..
I'm sure you know, if I is gonna be twice as big, T is gonna be twice as big..
So exactly how much is T going to increase, if I increases by 0.3% ?

Answer 34

I got that t is now 8.942 x 10^33

Answer 35

@jknighton1 - I don't know, I didn't calculate it. But why do you insist on plugging numbers in there? I don't think, they want some number as a solution..

Think about the last thing I posted for a sec..

Answer 36

T would increase by 0.3

Answer 37

Not by 0.3, but by 0.3% (I'm sure, you meant that) - EXACTLY!

e voilà - there you have it... If earth's rotational inertia would increase by 0.3%, so would the length of it's days.. another way to say this, is: Earth's rotational inertia is directly proportional to the length of it's days. \[T \propto I\]