How fast would the Earth need to spin for us to feel weightless? It's a classic question with many answers all over the Internet, but none here so I figured I'd ask it:
How fast would the Earth need to spin for a person (or anything for that matter) to feel weightless while on the surface at the equator?  
In this situation everything on the Earth's surface would essentially be in orbit around the Earth at the radius of the Earth's surface (let us assume the atmosphere was also spun up to this angular velocity so there would be no air drag slowing things down).  Let us also say by "surface of the Earth" we mean mean sea level.
You can decide for yourself if/how to factor in the bulge of the Earth.  You can assume that the Earth somehow is able to maintain its present shape while spinning up.
Any comments on whether an Earth spinning slightly faster than this speed will cause it to break apart or not will also be appreciated.
 A: How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force?
If you do the math (equating $G M m / R^2$ to $m \omega^2 R$ and using $M = \frac{4\pi}{3} \rho R^3$ as well as $\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the density $\rho$ of the sphere enters into the equation for $f$, the number of revolutions per unit time: $$f^2 = \frac{1}{3\pi}G\rho$$ 
For $\rho = 5.5 \times 10^3$ kilogram per cubic meter (the density of planet earth) it follows that $f=0.197 \times 10^{-3}$ revolutions per second, corresponding to a revolution period of $5070$ seconds (1 hour and 24 minutes).
A: This is an interesting question. Unfortunately it's impossible for everyone to be weightless at the same time because the spin effect would mostly effect people living on the equator and people on the poles would practically not notice it.
Also if it's at the sea level then seas would be weightless at the surface as well and eventually that water would fly off. Not to mention everything above the sea level, that is not strongly bound to earth, would fly off right away, meaning you would need to be tied to something in order to not become an astronaut if you jump (even a bit, even if you would just walk).
Oh yea, if you would spin earth that fast eventually it would stretch on the equators  and become a disc and probably totally fall apart...
A: It is not necessary to assume that the Earth has constant density, only that it retains the same shape when the rotation rate is increased. This is of course not realistic, because the Earth's crust rests on a thick fluid mantle which would deform if the forces acting on it changed.
According to wikipedia the effective strength of gravity at Earth's equator is $g_e=9.780m/s^2$, and where the radius is $r=6378km$. The rotation rate is $\omega_0=7.292\times 10^{-5}\text{rad/s}$. 
If the effect of gravitational attraction alone is $g$ and the rotation rate of the Earth is $\omega$ then $g_e=g-r\omega^2$. In order to nullify effective gravity $g_e$ at the equator we would need a new rotation rate $\omega_1$ for which $g_e=0$ hence $g=r\omega_1^2$. Substituting the current value of $g_e=g-r\omega_0^2$ we get $$ (\frac{\omega_1}{\omega_0})^2=\frac{g_e}{r\omega_0^2}+1$$ 
Using the above figures we get $\frac{\omega_1}{\omega_0}\approx 17$. The Earth would have to spin 17 times faster before people at the Equator became weightless.  
A: In view of the difference in orders of magnitude of previous answers, let’s see what dimensional analysis can tell us.
We look for a dimensionless combination ${\varpi}$ of $v$, $M$, $G$ and $g$, where $v$ is the velocity, $M$ is the mass of the Earth, $G$ the gravitational constant and $g=9.8m/s^2$ the gravitational acceleration at the surface of the Earth. By elementary manipulations we get that 
$$
\varpi=v \left(GM g \right)^{-1/4}
$$
is dimensionless. Thus, 
$$
v=R\omega = (GM g)^{1/4}\quad \Rightarrow \quad
\omega =\frac{(GMg)^{1/4}}{R}
$$
with $R$ the radius of the Earth.  Plugging numbers we get
$$
\omega\sim 1.2\times 10^{-3} \hbox{sec}^{-1}\, ,
$$
which would agree almost exactly with the answer of @sammy gerbil.
A: We already know gravity at equator, $mg$.
Centripetal acceleration equals gravitational acceleration $\implies$ weightless.
$$\omega^2R = \omega^2(6,400,000 m)=9.8\frac{m^2}{s}$$
$$\implies \omega = 0.00124 \frac{rad}{s} = 0.0118 rpm $$
Or 17 revolutions per day, ($\mathbf{\tau = 5070s}$), every 1 hour and 24 minutes.
Matches the checked answer. The changing shape of earth will matter if we get further from $cog$.
Btw you weigh a pound less flying West for the same reason: Is a westward flying plane heavier than an eastward one?
