If everyone on the earth lined up around the equator and started running east at 2.5 m/s (EDIT: velocity is relative to surface of the Earth).
How much does the length of a day increase?
Assume there's \(7x10^9\) people each with an average mass of \(m_p=55\) kg. Also, assume the earth to be a solid homogeneous sphere. mass of earth: \(m_e=5.97x10^24\) kg
radius of earth: \(r_e=6.37^6\) m
I understand this is a conservation of angular momentum problem, but I can't seem to get the numbers to work with the answer.
Could someone give me some guidance?

- PhoenixFire

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- theEric

If \(m_\text{everyone}\) is the mass of the people, then \(m_\text{everyone}=7\times 10^9\times 55\ [\text{kg}]\)
So the momentum before hand is the mass of the Earth put into the momentum of a sphere formula \(\left(\dfrac{2}{5}m_er_e^2\right)\omega\), plus the momentum of everybody put into the momentum of a hoop \((m_\text{everyone}r_e^2)\omega\) where \(\omega_1 = 0\).
I chose to say that the world is still, so that the difference in Earth's speed is the difference from normal.
I didn't finish the rest, though.

- theEric

So \(\omega_\text{everyone running}=(2.5\ [\text m/\text s])(r_e)-\omega_e\) where \(\omega_e\) is Earth's angular velocity.
And, of course, \(\omega_e=\omega_e\).
So, after the people start running, you have \(L_\text{after}=\left(\dfrac{2}{5}m_er_e^2\right)\omega_e+(m_\text{everyone}r_e^2)\omega_\text{everyone running}\)
If that is right.. Then it would be helpful to solve for \(\omega_e\)...

- anonymous

This is the best problem I have ever seen. Ever.

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## More answers

- theEric

Haha!

- theEric

\(L_\text{after}=\left(\dfrac{2}{5}m_er_e^2\right)\omega_e+(m_\text{everyone}r_e^2)(2.5\ [\text m/\text s])(r_e)-\omega_e\\
=\left(\dfrac{2}{5}m_er_e^2\right)\omega_e+m_\text{everyone}r_e^3(2.5\ [\text m/\text s])-m_\text{everyone}r_e^2(2.5\ [\text m/\text s])\omega_e\\
=....
\)
I should probably work on my homework now, though...

- anonymous

I disagree with the initial setup though - I think that the initial angular momentum would be the mass of the earth plus the mass of all the people on earth traveling at the rate of the earths spin.
\[L_i = I_e\omega_e =\frac{2}{5}M \ r_e^2 \ \omega_e = \frac{2}{5}\Big(m_{_{earth}}m_{_{everyone}}\Big)r_e^2 \ \omega_e\]

- anonymous

assuming the people are completely evenly distributed... :P

- theEric

Okay! I think its all relative. So I chose the normal situation to be \(\omega_0\) and then \(\omega_e\) would be the difference. But you can find Earth's new angular velocity, and subtract it from the normal! I think they would both yield the same result, with maybe some difference due to rounding differently.

- theEric

@AllTehMaffs
\[L_i = I_e\omega_e =\frac{2}{5}M \ r_e^2 \ \omega_e = \frac{2}{5}\Big(m_{_{earth}}m_{_{everyone}}\Big)r_e^2 \ \omega_e\]
I don't know if this is right because it looks like it says that the Earth and everyone have the moment of inertia of a solid sphere, which isn't the case. The people are lined up around the equator, which is \(r_e\) perpendicularly from the axis of rotation. Is that right?

- anonymous

I definitely misread your original post, sorry. My defense in saying they're equally spread about the earth is that it kinda says that they only line up when they start running - at least that's how I read it :P I think your way might be more right though.
But in any case shouldn't the angular velocity of everyone be
\[\omega_{everyone} = \frac{2.5m/s}{r_e} + \omega_e\]
the angular velocity is the velocity divided by r, and it would be added, not subtracted to the new rotation of the earth because they're moving in the same direction as it's spinning.
?

- theEric

Haha, thanks. I did mess up on \(\omega_\text{everyone}\).
It should be \(\dfrac{v_\text{people}}{r_e}\) instead of multiplication. I chose subtraction because the people run that fast relative to the earth, which has it's own angular velocity in the \(\sf opposite\) direction to that of the runners.|dw:1383547381788:dw| I think that makes sense, but I'm not quite 100% confident in that.

- anonymous

Methinks that the earth spins west to east, and the runners are running east
|dw:1383547704396:dw|

- theEric

Yep! That scenario is a little more complicated than mine, where I chose \(\omega_0=0\) as the initial angular acceleration. So, your picture makes sense when looking from outerspace, and mine makes sense if you are rotating with \(\omega_0\).

- anonymous

ooooooo, I get you

- anonymous

^_^

- theEric

:)

- PhoenixFire

Sorry I've been quite for all this. I had to go to training and just got home now.
So I my thoughts on some of the above:
1) Are you allowed to assume that the rotation of earth is 0? @theEric Because wouldn't that be counter to what the problem is trying to achieve? as we're trying to slow down the rotation of earth, thus increasing the length of the day.
-- I just reread what you wrote, and by assuming the perspective of a person on the planet I think it might be safe to take \(w_e=0\) however, I still think my comment has some merit and needs explaining? (possibly due to point (3)?)
There's so much in this i forgot to think about hahaha.
2) I think you're right to assume that the initial angular moment of earth is inclusive of the mass of everyone on the planet too. (Forgot to do this in my calculations) you'd have to take the moment of inertia of the earth as a solid sphere plus the moment of inertia of the "hoop" of people surrounding the equator.
3) the angular velocity of everyone \(w_{everyone}=\frac{v}{r_e} + w_{earth}\) would be wrong as I forgot to mention that the velocity of 2.5 m/s is relative to the surface of the earth. would this not entail that 2.5m/s is inclusive of the earth current rotation?

- anonymous

I can answer 3)
@theEric doesn't use that definition in his calculations, so I can't speak for his way (cause I keep messing his way up) but that definition is from my way and is correct when you set it up without making anything zero at the beginning.
Like you said, the velocity is"relative to the surface of the earth," and the surface of the earth is spinning at some angular velocity already so by definition they would be moving at an angular velocity whose total velocity is their running velocity *plus the velocity of the surface they're running on.
^_^

- PhoenixFire

Ah yes! That would make sense, because you've taken a perspective from off the planet, right?
I have the final answer, but I still can't make my numbers work out.
Length of day increases by \(7.5\times 10^{-11}\) seconds.

- anonymous

That change in day was totally worth the effort from all of the people ^_^ We can try my method if you want, or wait for theEric's fanciness ^_^
How'd you set up your problem?

- anonymous

And yeah, the perspective of my setup is frrooooommm spppaaaaace.

- PhoenixFire

I tried a bit of brute force. Just equating the angular momentum and solving for \(w_e\)
You have \(L_e=I_ew_e\) and \(L_{people}=I_{people}w_{people}\)
But because \(w_{people}=\frac{v}{r_e}\) we can simplify \(L_{people}\)
\[L_{people}=I_{people}\frac{v}{r_e}=m_{people}r_{e}^2\frac{v}{r_e}=m_{people}r_ev\]
So equating these two: \(\large I_ew_e=m_{people}r_ev\) and solving for \(w_e\) But this gives a number many orders of magnitude larger than what it's supposed to be.
Taking \(I_e=\frac{2}{5}m_er_e^2\) and \(v=2.5\) m/s
This I believe is also taking perspective of someone on the planet.
I noticed that the mass of the people when calculating \(I_e\) is negligible too.

- PhoenixFire

Even if you taking \(\large L_{people}=I_{people}(\frac{v}{r_e}+w_e)\) it still doesn't work.

- anonymous

Browsing a bit of XKCD, I found something very very similar.
http://what-if.xkcd.com/41/

- anonymous

Well let's start from what we absolutely know first then ^_^ Conservation of angular momentum.
\[L_i = L_f\]
\[L_i = L_{Earth \ 1} + L_{people \ 1} = I_E\omega_1 + i_P\omega_1
\\ \
\\ L_f = L_{Earth \ 2}+ L_{people \ 2}\] = I_E\omega_2 + i_P(\omega_2+\omega_P)
Initial
\[L_{Earth \ 1} = \frac{2}{5}m_Er_E^2\omega_1
\\ \
\\w_1 = 1rev/day
\\ \
\\ \
\\ L_{people \ 1} = m_Pr_E^2\omega_1\]
final
\[L_{Earth \ 2} = \frac{2}{5}m_Er_E^2\omega_2
\\ \
\\ L_{people \ 1} = m_Pr_E^2(\omega_2+\omega_P)
\\ \
\\ \omega_P = (2.5m/s)/r_E\]
\[\frac{2}{5}m_Er_E^2\omega_1 + m_Pr_E^2\omega_1 = \frac{2}{5}m_Er_E^2\omega_2 + m_Pr_E^2(\omega_2+\omega_P) \]
we have everything but omega 2

- PhoenixFire

Yeah I just did that, and solved for \(w_2\), the change doesn't recognize on my calculator. it still works out to be the exact same rotational period as earth has normally. unless I messed up in the rearrangement.

- anonymous

\[ \frac{2}{5}m_Er_E^2\omega_1 + m_Pr_E^2\omega_1 - m_Pr_E^2\omega_P=\Big( \frac{2}{5}m_Er_E^2 + m_Pr_E^2 \Big)\omega_2
\]
\[ \omega_2 = \frac{ \frac{2}{5}m_Er_E^2\omega_1 + m_Pr_E^2\omega_1 - m_Pr_E^2\omega_P}{\frac{2}{5}m_Er_E^2 + m_Pr_E^2 }
\]
\[=\frac{ \cancel{ r_E^2}\Big(\frac{2}{5}m_E\omega_1 + m_P\omega_1 - m_P\omega_P\Big)}{\cancel{r_E^2} \Big(\frac{2}{5}m_E + m_P \Big) }
\]
what did you use as omega 1? It has to be in meters per second for the units to work out right
\[ \omega_1 = 1rev/day = \frac{2 \pi r_E}{86400 s} \]

- anonymous

and mp is the mass of everyone and their mother, not just the 55kg

- anonymous

cheese and crackers I'm a dumb. That's not the angular velocity at all. Sorry :P

- PhoenixFire

I don't see why it needs to be meters/second. radians per second should work too right?

- anonymous

yeah, I just blanked. just 2pi / 86400

- anonymous

sorry about that

- PhoenixFire

Yup, that's what I used. but \(w_2\) ended up being the same as \(w_1\)

- anonymous

hmph. Well it's going to be a ridiculously small number...

- anonymous

\[= \frac{2m_E\omega_1/5 + m_P\omega_1−m_P\omega_P)}{(2m_E/5+m_P)}\]
\[=\cancel{\frac{2m_E/5+m_P}{2m_E/5+m_P}}\omega_1 - \frac{m_P \omega_P}{2m_E/5+m_P}\]
\[ =\omega_1 - \frac{m_P \omega_P}{\frac{2}{5}m_E+m_P}\]

- anonymous

you can just use that little bit as the change

- anonymous

just convert
\[- \frac{m_P \omega_P}{\frac{2}{5}m_E+m_P}\]
to the right units
It's in rad/s , just figure out how many seconds that is for 1 revolution

- anonymous

then you throw away all the giganto terms that are throwing everything off

- PhoenixFire

Same number i've been getting all along.
\(6.3274 \times 10^{-20}\) Ridiculously small.
This is in rad/s.. so I use the formula \(w=2\pi f\) and \(f=\frac{1}{T}\) where T is the period.
so you get \(T=\frac{2\pi}{w}\) which just works out to be 86400 seconds.
We must be doing something wrong.... somewhere.

- anonymous

excuse me.. but when 7 billion people line up on the equation.. doesn't that alone change the length of the day? shouldn't we be considering that as well?

- anonymous

equator

- anonymous

sorry.. i guess that ll hardly matter!

- anonymous

@Mashy They're completely evenly distributed and only move north to south so the net force is zero. Totally legit. ^_^
Also, theEric's equation gives exactly the same thing, so I'm pretty sure the answer is right, we're just not using it correctly.

- anonymous

and there's a pre-built bridge across the Atlantic and Pacific....

- anonymous

no wait.. what i meant is.. hwen people shift their positions, the moment of inertia increases right? cause they are all coming towards the equator?

- anonymous

Yeah, very much so, all of the "r" vectors pointing to each person that you'd sum over for the moment of inertia of the mass of humans would go craaaazy.

- anonymous

it doesn't actually go crazy..compared to mass of the earth :P

- anonymous

well there's that little fact.... ^^

- theEric

Hi! I've skimmed through the new responses. Is there a consensus on an answer or method?
@mashy - you have a good point - just the fact that everyone's on the equator changes the Earth-people moment of inertia! If you wanted to be more realistic, it would be a great idea to assume the people to be dispersed evenly across the Earth as a \(\sf shell\). I know there are more people closer to the equator. Defining a moment of inertia for that is probably beyond what is expected for this problem. And then the actual population densities around the world would be very difficult to account for. That would be beyond a conceptual physics problem like this. But a shell might not be a bad idea. It's one way!
The change in the day, though I haven't calculated it or came to a definite formula, would be small! The moment of inertia of the Earth (solid sphere, lots of rock and stuff) is much greater than the moment of inertia of the ring of people around the equator. All humans are light compared to the surface of the Earth, and the Earth is so much larger than what we have at the surface. So humans walking probably couldn't do a whole lot!

- anonymous

The two methods that use humans as a ring around the earth initially produce the same result
Straight forward total conservation of momentum that sets nothing to zero yields
\[\omega_2 = \omega_1 - \frac {m_P\omega_P}{\frac{2}{5}m_E + m_P} \]
and setting angular velocity as initially zero yields the same expression minus the omega 1
\[\omega_2 = - \frac {m_P\omega_P}{\frac{2}{5}m_E + m_P}\]
Neither of which give useable results, although I'm completely blanking on how to use the given angular velocity to find the change in time in 1 day :P

- PhoenixFire

@theEric @AllTehMaffs
I figured out how to do it. My lecturer helped a bit.
The earth itself has an initial angular momentum \(L_{earth}=I_e\omega =\frac{2}{5}m_er_e^2\omega \).
All the people lined up will have an angular momentum of \(L_{people}=I_p\omega=Nmrv\) where N is the number of people. Now because they have this angular moment the earth has to lose this much moment to conserve total momentum.
So the fraction that \(L_{people}\) is of \(L_{earth}\) is the same fractional amount that is lost in the time of a day.
\[\Delta T=\frac{L_{people}}{L_{earth}} T_{day}\]

- theEric

Interesting! Thanks for sharing, @PhoenixFire !

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