If the ground's normal force cancels gravity, how does a person keep rotating with the Earth? When I am on earth, the weight of my body is countered by the reaction of the ground. So, there is no net force acting on me.
But I am spinning with earth. But if there is no centripetal force then why am I spinning? And the equal air pressure on both side of my body won't be enough for me to stay in the same angular velocity as the earth.
Is it just conservation of angular momentum?
 A: You may underestimate how small the effects of the Earth spinning are.  While they have profound effects on large scales, on the scale of your body, they are small.
You can calculate the centripetal acceleration required to stay on the Earth's surface.  Even near the equator, where the effect is strongest, you're still talking about around 0.03m/s^2.  Its hard for you to detect an acceleration that slight.  That acceleration becomes more important when we make devices which are more sensitive, such as pendulums designed to show the rotation of the Earth.
A: Actually, this is rather insightful. The normal force from the ground does not quite cancel out the effect of gravity. The difference between them is precisely the centripetal force that keeps you rotating around with the Earth's surface.
Of course, you won't notice this because the centripetal force is so small compared to the gravitational force on you. The centripetal acceleration at the equator is
$$a_c = \omega^2 r \approx \biggl(\frac{2\pi}{24\ \mathrm{h}}\biggr)^2\times 3959\text{ miles} = 0.034\ \frac{\mathrm{m}}{\mathrm{s}^2}$$
which is a paltry one-third of a percent of the gravitational acceleration, and at higher latitudes it is correspondingly less.
A: It is easier to consider you standing on the Equator.
Assume that the gravitational field strength at the Equator is $g$.  This would be the acceleration of free fall at the Equator with no air resistance if the Earth was not spinning.  
If the reaction of the Earth is $N$ then assuming down is positive and using N2L, $mg-N=0$ if your mass is $m$.
If the Earth of radius $R$ is spinning with angular speed $\omega$ then using N2L one gets $mg-N'=mR\omega^2$.
So the reaction force due to the Earth $N'$ has decreased.
The acceleration of free fall would also decrease to $g-R\omega^2\;(\approx 0.03 \rm ms^{-2})$ as would your apparent weight $m(g-R\omega^2)$.
So measuring your "weight" at the Equator using a spring balance would yield a smaller value than that at the geographic poles where you your centripetal acceleration would be zero.
If it so happened that the period of rotation of the Earth was 84.5 minutes you would find that there was no reaction force due to the Earth and the acceleration of free fall would be zero.
Objects which you let go of would not fall closer to the Earth.
This would be a state of weightlessness.
It so happens that 84.5 minutes is the theoretical speed of a satellite of the Earth whose circular orbit had a radius equal to that of the Earth.
All this has ignored the effect of air resistance and the fact that if the Earth was made to spin that fast it would disintegrate due to the brittle crust not being very good at sustaining tensile stresses.
A: Say you stand on a scale on the surface of Earth, and that it shows your weight $\vec{W}=m\vec{g}$. It is precisely balanced by the normal force $\vec{N}$. The local gravitational constant, little $g\approx 9.8 ~\mathrm{m/s^2}$, is not just due to gravity, despite the name. It is actually a vector sum $\vec{g}=\vec{g}_{gr}+\vec{g}_{cf}$ of gravitational acceleration $\vec{g}_{gr}$ and centrifugal acceleration $\vec{g}_{cf}$. Therefore, you have already implicitly accounted for the fictitious force, the centrifugal force $m\vec{g}_{cf}$, which is present in the accelerated reference system of you & the scale. 
A: Centripetal force is tricky and counter-intuitive. When I first took physics, it tripped me up because I kept trying to fit it into $F_{net} = ma$ in the wrong way. No matter how I tried to look at it, I couldn't come to any conclusion other than that centripetal force is actually directed away from the center.

In the beautiful diagram above, we have a ball on the end of a string being swung around a center point.
Baby Devsman did not understand. Baby Devsman reasoned that if there is a force $T$ supplied by the string, then the ball must be getting closer to the center point except for this mystical magical force $F_c$ which is pulling it away by virtue of it going in a circle. Then, $F_{net} = ma$ as satisfied as the ball stays a constant distance from the center point. Why did scientists say it was directed toward the center? It clearly couldn't be, or else the ball would move closer to the center really fast.
$$0 = T + F_c???$$
This is wrong! Baby Devsman had much difficulty understanding many concepts as a result of wrong thinking. Baby Devsman would later learn that centripetal force is not a thing. Electric force is a thing. Electric charges attract each other and a force pulls them together. Normal force is a thing. The ground pushes back against stuff that sits on it.
Centripetal force, though, is not a thing. Nothing exerts centripetal force. Moving in a circle does not cause a force to be exerted. Centripetal force is a requirement of circular motion. Now I understand that centripetal force does not oppose $T$ in the beautiful diagram, but that the required centripetal force to satisfy the assumed scenario is provided by $T$. The ball does not get closer to the center point, true, but the ball's circular motion requires a particular $F_c$. That is, if an object is going to move in a circle,
$$F_c = F_{net}$$
Which in the case above, means $F_c = T$.
What does this have to do with Earth?
When you stand on the ground, your assertion that the weight of your body is matched by the ground's normal force is only approximately true. In reality, the centripetal force required for your body to move around the Earth is satisfied by the net of your body weight and the ground's support force:
$$F_c = W - N$$
As DavidZ says, the centripetal force is small compared to your weight, as the Earth's rotation and radius result in a small centripetal acceleration compared to its gravitational acceleration, but it is still there. As far as what accelerates you to move alongside the Earth, friction does.
Friction is modeled by:
$$F_f = \mu N$$
Notice that the $N$ is normal force, not $F_{net}$. This same N that (almost) matches your weight.
Interestingly, this means that walking toward the equator is ever so slightly (not nearly noticeably) more difficult than walking away from it, as you're having to accelerate yourself to keep up with the change in the radius of the cross-section of the Earth at your changing latitude.
A: Why am I spinning?
To be precise, lets assume you are an astronaut that has returned to earth after being in a non-geosynchronous orbit.  After landing you are now at rest with respect to the surface of the earth in your general area.
The reason you are spinning with the earth is quite simple.  Gravity is pushing you against the surface of the earth.  There is friction between you an the earth as a result of the pressure between you and the surface.  That pressure is caused by your weight i.e. gravity.  That friction causes your body to move with the surface in the same way that a moving sidewalk moves you when you are standing on it.
But if there is no centripetal force then why am I spinning?
Centripetal force does not initiate your motion.  The speed component of your velocity is not a result of centripetal force.  Centripetal force changes the direction component of your velocity.  In other words centripetal force changes the direction of your movement such that you follow the curved path around the center of the earth.  Another way think about it is: the direction of your velocity is always orthogonal to the direction of the centripetal force and therefore the centripetal force doesn't contribute to the speed component of your velocity.
For reference, here's a simple diagram of centripetal motion and how the force and velocity relate (source):

Centripetal force, in this example, is always directed towards the center of the earth which, not coincidentally, is the same direction that gravity points.  The reason is that in this case the centripetal force is caused by gravity.  If gravity were to somehow stop pulling you towards the center of the earth your body would continue in a straight line tangent to the earth.  At first you would seem to float up but as you continued on this path, you would leave vicinity of the earth entirely.
A: You say

If there is no centripetal force

But that is not true. When you measure your weight at the North Pole or at the Equator, you get a different answer. The shape of the Earth (a slightly flattened sphere) is part of the reason*); but the rotation of the earth (which incidentally causes the flattening) also plays a role. At 24 hr/revolution, and a circumference C of 40,000 km, the acceleration is
$$a = \frac{v^2}{R} = \frac{\left(C/(24*3600)\right)^2}{C/2\pi}=\frac{2\pi\times 4\times 10^7}{24*24*3600*3600}\approx 0.034~ \rm{m/s^2}$$
This is described in more detail here.
When you are at a point away from the equator, the direction that you think of as "up" is not, in fact, exactly pointing away from the center of the Earth because there is again a lateral acceleration - so if gravity points to the center of the Earth, if you hang a mass on the end of a string it will in fact deviate from that line. The acceleration will scale with the cosine of the latitude - the lateral component of that force will scale with $\cos\lambda \sin \lambda$, which will reach a maximum at 45° latitude. At this point, your sense of "up" will be off by about 0.017/9.8 radians, or about 0.1°

In this diagram, the red vector is the force of gravity - which, if you are "leaning in" slightly, becomes a centripetal force (blue) and the apparent force of gravity (green). It's not much - but just enough to keep all of us doing our circles with the planet.

*)The other part of the reason is that the oblate shape of the Earth means you are a little closer to the center of mass at the North Pole; the radius at the equator is 6378 km, while it's 6357 km at the poles. If all mass of the earth were compressed into a smaller sphere, this would result in a difference in gravity of about 0.67% (heavier at the poles); but when you take into account the mass distribution (mass near the equatorial bulge has relatively little effect on gravity at the poles) it's a bit less - about 0.2%; see link above, and this interesting question and associated answers. 
