# Where does the component of the centripetal force in circular motion come from? [duplicate]

If a man is standing on the equator of a non-rotating planet then the normal force is equal and opposite to the weight force.

If the planet now starts rotating then there would be a net (centripetal) force directed towards the centre equal to $mg-N$. If the planet continued to increase its rotational speed then eventually the centripetal force would equal the weight force and the man would be in orbit (on the surface of the planet). My question is where does this net force stem from? Is the centripetal force a part of the weight force or is the centripetal force opposing and reducing the normal force? If it is opposing and reducing the normal force then wouldn't this require an actual force to do so?

The centripetal force is the total sum of all forces along the radial direction. The weight minus the normal force in this case.

Weight and normal force are usually set equal in equations, yes, but in fact the weight is a tiny bit larger. That tiny bit is what constitutes the centripetal force.

First, formulate the Balance equation for the Forces as follows:

Introduce a coordinate System which (for simlpicity) are polar coordinates $\vec{x} = (x,y) = (rcos \phi,r sin\phi)$. You have to consider the Balance of Forces in the $r$-direction for the man. It is given by

$m\ddot{\vec{x}}\vec{e_r} = N-mg$

where $m$ is the mass, $g$ gravity acceleration, $N$ the normal force (acts only if the man is in contact with planet) and $\vec{e}_r$ the unit vector in $r$-direction. This unit vector Points in the same direction as $\vec{x}$ is pointing (radius Points from origin to the Position of the man); you have

$\vec{e_r} = \frac{\vec{x}}{|\vec{x}|}$.

After differentiating the Position vector, you obtain:

$m\dot{\vec{x}} = (\dot{r}cos \phi - \dot{\phi}rsin \phi,\dot{r}sin \phi + \dot{\phi}rcos \phi)$

$m\ddot{\vec{x}} = (\ddot{r}cos \phi - 2\dot{\phi}\dot {r}sin \phi -\ddot{\phi}rsin \phi -r\dot{\phi}^2cos \phi,\dots)$

After simplification you get

$m\ddot{\vec{x}}\vec{e_r} = m \ddot{r}-mr \dot{\phi}^2 = N-mg$.

If the man does not move radially (Equilibrium condition) you have $\ddot{r} = 0$, hence:

$N = m(g-r \dot{\phi}^2)$.

The normal force from earth surface on man decreases if the force term $mr \dot{\phi}^2$ increases; therefore this term is a centrifugal force term. The relation

$m \ddot{r} = mr \dot{\phi}^2 + F$

for some other external Forces $F$ Shows, that the centrifugal term causes an acceleration in positive r-direction, i.e. from the planet away.