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 one year ago
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?
 one year ago
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?

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theEric
 one year ago
Best ResponseYou've already chosen the best response.1If \(m_\text{everyone}\) 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
 one year ago
Best ResponseYou've already chosen the best response.1So \(\omega_\text{everyone running}=(2.5\ [\text m/\text s])(r_e)\omega_e\) 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 \(\omega_e\)...

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1This is the best problem I have ever seen. Ever.

theEric
 one year ago
Best ResponseYou've already chosen the best response.1\(L_\text{after}=\left(\dfrac{2}{5}m_er_e^2\right)\omega_e+(m_\text{everyone}r_e^2)(2.5\ [\text m/\text s])(r_e)\omega_e\\ =\left(\dfrac{2}{5}m_er_e^2\right)\omega_e+m_\text{everyone}r_e^3(2.5\ [\text m/\text s])m_\text{everyone}r_e^2(2.5\ [\text m/\text s])\omega_e\\ =.... \) I should probably work on my homework now, though...

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1I 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. \[L_i = I_e\omega_e =\frac{2}{5}M \ r_e^2 \ \omega_e = \frac{2}{5}\Big(m_{_{earth}}m_{_{everyone}}\Big)r_e^2 \ \omega_e\]

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1assuming the people are completely evenly distributed... :P

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Okay! 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
 one year ago
Best ResponseYou've already chosen the best response.1@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?

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Haha, 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.

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1Methinks that the earth spins west to east, and the runners are running east dw:1383547704396:dw

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Yep! 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 \(\omega_0\).

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1ooooooo, I get you

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1Sorry 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?

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Ah 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.

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1That 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?

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1And yeah, the perspective of my setup is frrooooommm spppaaaaace.

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Even if you taking \(\large L_{people}=I_{people}(\frac{v}{r_e}+w_e)\) it still doesn't work.

Lessis
 one year ago
Best ResponseYou've already chosen the best response.0Browsing a bit of XKCD, I found something very very similar. http://whatif.xkcd.com/41/

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1Well 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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.

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{2}{5}m_Er_E^2\omega_1 + m_Pr_E^2\omega_1  m_Pr_E^2\omega_P=\Big( \frac{2}{5}m_Er_E^2 + m_Pr_E^2 \Big)\omega_2 \] \[ \omega_2 = \frac{ \frac{2}{5}m_Er_E^2\omega_1 + m_Pr_E^2\omega_1  m_Pr_E^2\omega_P}{\frac{2}{5}m_Er_E^2 + m_Pr_E^2 } \] \[=\frac{ \cancel{ r_E^2}\Big(\frac{2}{5}m_E\omega_1 + m_P\omega_1  m_P\omega_P\Big)}{\cancel{r_E^2} \Big(\frac{2}{5}m_E + m_P \Big) } \] what did you use as omega 1? It has to be in meters per second for the units to work out right \[ \omega_1 = 1rev/day = \frac{2 \pi r_E}{86400 s} \]

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1and mp is the mass of everyone and their mother, not just the 55kg

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1cheese and crackers I'm a dumb. That's not the angular velocity at all. Sorry :P

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1I don't see why it needs to be meters/second. radians per second should work too right?

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1yeah, I just blanked. just 2pi / 86400

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1Yup, that's what I used. but \(w_2\) ended up being the same as \(w_1\)

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1hmph. Well it's going to be a ridiculously small number...

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1\[= \frac{2m_E\omega_1/5 + m_P\omega_1−m_P\omega_P)}{(2m_E/5+m_P)}\] \[=\cancel{\frac{2m_E/5+m_P}{2m_E/5+m_P}}\omega_1  \frac{m_P \omega_P}{2m_E/5+m_P}\] \[ =\omega_1  \frac{m_P \omega_P}{\frac{2}{5}m_E+m_P}\]

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1you can just use that little bit as the change

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1just 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

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1then you throw away all the giganto terms that are throwing everything off

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1Same 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.

Mashy
 one year ago
Best ResponseYou've already chosen the best response.0excuse 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?

Mashy
 one year ago
Best ResponseYou've already chosen the best response.0sorry.. i guess that ll hardly matter!

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1@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.

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1and there's a prebuilt bridge across the Atlantic and Pacific....

Mashy
 one year ago
Best ResponseYou've already chosen the best response.0no wait.. what i meant is.. hwen people shift their positions, the moment of inertia increases right? cause they are all coming towards the equator?

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, 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.

Mashy
 one year ago
Best ResponseYou've already chosen the best response.0it doesn't actually go crazy..compared to mass of the earth :P

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1well there's that little fact.... ^^

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Hi! 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 Earthpeople 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!

AllTehMaffs
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1Interesting! Thanks for sharing, @PhoenixFire !
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