# Homework Help: Conservation of angular momentum

1. Dec 10, 2009

### scoldham

Hi, Long time pf reader... first time poster in need of physics help.

1. The problem statement, all variables and given/known data

The moment of inertia of the Earth is approximately 0.331MERE2. If an asteroid of mass 5.0 × 1018 kg moving at 150 km/s struck (and stuck in) the Earth’s surface, by how long would the length of the day change? Assume the steroid was traveling westward in the equatorial plane and struck the Earth’s surface at 45◦.

$$\omega$$E = 1/86400 rev/sec

Sub A is for Asteriod
Sub E is for Earth
Sub sys is for system (Asteroid + Earth)

2. Relevant equations

Isys $$\omega$$sys = IE $$\omega$$E + VA

Note: VA = RA $$\omega$$A

Isys = (0.311*ME -MA) RE2

3. The attempt at a solution

So essentially, I derived the equations above based on conservation of angular momentum. I assume the point of the line in the problem statement about 45 degrees striking along the equatorial plane is telling me the asteroid's momentum is exactly converse the earth's ergo 1 dimension.

After calculating Isys, I solved the first equation for $$\omega$$sys and then plugged in the given data... I'm getting 1.16 * 10-5 for $$\omega$$E implying (to me at least) that the change in the length of a day is less than 100 seconds if that...

I would think that even though the asteroid is 6 times smaller than the Earth, it would make more of a difference than that... am I right? Guidance much appreciated.

Last edited: Dec 10, 2009
2. Dec 10, 2009

### kuruman

If you are going to use Iω for angular momentum, you need to express ω in rad/s not rev/sec

What is the angular momentum of the asteroid with respect to the center of the Earth just before the collision?

Irrelevant.

The mass of the Earth is 6x1024 kg, the asteroid's mass is much less than one-sixth of that. Perhaps you meant six orders of magnitude smaller? It's not the same as "six times smaller."

3. Dec 11, 2009

### scoldham

Thanks for the response. Looks like I was right about my gut feeling of something being wrong.

Concerning the angular momentum of the asteroid w.r.t. earth, am I right in thinking that I can calculate this based on the following:

L = r $$\times$$ p

with
r = position vector [1 cos(45◦)]
p = linear momentum of the asteroid (MA VA)

I'm not super sure about the position vector....

At any rate, instead of adding VA to the initial equation I provided, I would add L?

Thanks again,

-Stosh

4. Dec 11, 2009

### dr_k

scoldham,

You should be headed for this result,

$$\rm \omega_f = \frac{I_E\omega_i-mvR_Esin\theta}{I_E+mR_E^2}$$

but first, you need to draw a nice picture of the collision, a before (initial) and after (final) set of pictures: (i) and (f).

Last edited: Dec 11, 2009
5. Dec 11, 2009

### scoldham

I'm guessing the m and v in the numerator are for the asteroid.. but I'm not sure about the m downstairs, is that mass of the earth?

I have some penciled sketches, which do help some. They have helped me to see why you would choose sin$$\theta$$ instead of cos$$\theta$$. One wants the distance between the cm of each object which would be relative to the striking angle.

6. Dec 11, 2009

### dr_k

Yes, m and v are for the asteroid.

The m downstairs is for the asteroid ( m << M_E)

Let me show you the picture I have:

http://img709.imageshack.us/img709/4576/am1.jpg [Broken]

There was a typo in my original equation, which I've now changed. I_e, should have been I_E:

$$\rm \omega_f = \frac{I_E\omega_i-mvR_Esin\theta}{I_E+mR_E^2}$$

This equation represents $${\vec L}^{\rm sys}_f = {\vec L}^{\rm sys}_i$$

$$I_E$$ is the moment of inertia of the Earth. The $$mR_E^2$$ term is the moment of inertia for the asteroid once it's embedded into the Earth.

Last edited by a moderator: May 4, 2017
7. Dec 11, 2009

### scoldham

I think I have finally got it!

Using the picture it is easy to see the triangle formed between where the asteroid starts, where it strikes and CME. This gives us

$$L_A = r_{AE} \times p_A = M_A V_A R_E sin \theta$$

which is the angular momentum of the Asteroid w.r.t. CME.

The angular momentum of the Earth is derived from

$$L_E = I_E \omega_E$$

(making sure that $$\omega_E$$ is in rads as pointed out earlier)

Conservation of angular momentum tells us

LFinal = LE Initial + LA Initial

with LFinal = IFinal $$\omega$$ Final.

Noting, IFinal = (.311 ME + MA) RE2 = IE + MA RE2

Substituting in for all the L's

$$I_E + M_A R_E^2 \omega_{Final} = M_A V_A R_E sin\theta + I_E \omega_{E}$$

Then solving for $$\omega_{Final}$$:

$$\rm \omega_{final} = \frac{I_E\omega_{E}+M_A V_AR_Esin\theta}{I_E+M_AR_E^2}$$

Which tells us the new angular velocity of the Earth after the collision which tells us the change in the time of a day.

Note: These equations are general.... it assumes that LA is a negative value (for this case at least).

Last edited: Dec 11, 2009
8. Dec 11, 2009

### D H

Staff Emeritus
Look at the units. Angular momentum has units of mass*length2/time. Your middle term has the correct units, but your final term does not. It has units of mass*length/time, i.e., linear momentum, not angular momentum.

9. Dec 11, 2009

### scoldham

Ah... I forgot to factor in $$R_E$$... the missing length factor. It is part of the position vector. Good catch.

It made it to the final solution. But it needs to be woven throughout the reasoning I suppose. I'll make edits.

10. Dec 11, 2009

### dr_k

There's quite a few typos in here, so you need to go over your calculations carefully. I'm not sure if they were typed in incorrectly (very easy to do!), or if they're incorrect in your notes, but yes, you're getting there!

Maybe this will help...

Looking at $${\vec L}^{\rm sys}_f = {\vec L}^{\rm sys}_i$$, implies

$$\rm \omega_f (I_E+mR_E^2) = I_E\omega_i-mvR_Esin\theta$$

Look at my original picture. You must be careful to get the (+/-) signs correct, for your pictures. Use the right hand rule to determine whether $$\omega$$ is into or out of the page. In this particular problem, it can be a little confusing. My picture show a view from the top, looking down on the N pole, where the Earth is rotating CCW. You're almost there...

n.b. You don't need to be concerned with the CM for the system, since $$\frac{m}{M_E} <<1$$. Just choose your inertial ref frame at the CM of the Earth.

11. Dec 11, 2009

### scoldham

I did state at the end of the reasoning that I'm assuming LA is negative, by virtue of the same reason.... If LA was hitting in the same direction of LE then it would be positive. Perhaps the way I went about stating it was incorrect, but I did realize that LA was to be subtracted in the actual numerical calculation.

Yes, I understand this. I was only using CME as a reference point in the picture... it was easier for me to see it looking at points as opposed to large masses =)

So, barring these points, and the case of the missing RE from earlier, I believe I have it, no?

12. Dec 11, 2009

### dr_k

Ahh, I didn't see the part where you stated you're going to put $$L_A$$ to be negative. So, yes, provided you put in LA as a negative number...but, and here comes the "but"! :)

When writing down variables, one must be very careful to represent them so they don't introduce sign errors. For example, if I write v, I'll always know that $$v = || \bf v||\ge 0$$, a magnitude, as opposed to a component, e.g. $$v_r$$, which could be negative. It's very easy to make a mistake with these sort of things. Just food for thought.

13. Dec 11, 2009

### scoldham

Thanks for all the help and advice! I think we can mark this one off as solved.

=D