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?

Question

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?

theeric · Answer

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 · Answer

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 · Answer

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

theeric · Answer

Haha!

theeric · Answer

$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 · Answer

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 · Answer

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

theeric · Answer

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 · Answer

@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 · Answer

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 · Answer

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 · Answer

Methinks that the earth spins west to east, and the runners are running east

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