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?
Answer
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.
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