What exactly does accelerometer measure on a vertical rotating disk? I am trying to understand an aspect of rotational dynamics. I am having some trouble arriving at a solution. Consider a disk of uniform mass distribution, and radius R centered at the origin of an xy coordinate system and rotating around the z by a motor with the constant frequency. 
I have figured out that the centripetal acceleration of the point located at the radius of $r$ is calculated as: 
$$a = \omega^2R$$
I was wondering if I mount an accelerometer on the surface of the disk, what does it measure?
Am I supposed to consider the gravity too? 
$$a_{total}= \sqrt{a^2+ g^2}$$
 A: I think that it may depend on your accelerometer. But today's MEMs (Microelectromechanical systems) accelerometers like the ones present on mobile phones does not sense gravity. It is if you let an accelerometer in free fall you must read 0 at the output. However, if the accelerometer is not in free fall, e.g. it is over a surface, the accelerometers will measure the acceleration of the reaction force, which for the case of a horizontal surface is, in fact, the gravity.
Note, however, that this vector will point in the opposite direction of the actual gravity's direction.
A: An accelerometer works by measuring the movement of a test mass - or the force required to prevent a test mass from moving. The measured acceleration is in the opposite direction to the movement of the test mass.
The raw data from an accelerometer at rest relative to the surface of the earth will show a $1$ g acceleration upwards. However, accelerometers are often zeroed to subtract the acceleration due to gravity. The zero point in a very sensitive accelerometer will have to be calibrated for a specific location since the effective value of g varies with latitude.
So if you are looking at raw acceleration data (called proper acceleration) and your accelerometer is fixed to a rotating disk then it will show the vector sum of the acceleration due to gravity and centripetal acceleration which is directed towards the centre of the disk.
A: An accelerometer measures the forces acting on its mount (the normal force) by means of 3 small beams that deflect. Since it has known masses of its part it can convert the forces into accelerations.
Nevertheless it just measures forces. So gravity will be measured only if the support resist the motion due to gravity (sitting on a tabletop). When free falling, no forces act on the mount and the accelerometer registers zero.
In your case let is figure out the forces needed to keep the accelerometer rotating in a circle under gravity

In the local orientation of xy the (kinematic) acceleration is
$$ \ddot{x} = - r \dot{\theta}^2 $$
$$ \ddot{y} = r \ddot{\theta} $$
and the forces acting are
$$ F_x -m g \sin \theta  = m \ddot{x} = - m  r \dot{\theta}^2 $$
$$ F_y - m g \cos \theta  = m \ddot{y} = m r \ddot{\theta} $$
and if we call $a_x  = F_x/m$ and $a_y = F_y/m$, and $\omega = \dot \theta$ then you will get
$$ a_x = g \sin \theta - r \omega^2 $$
$$ a_y = g \cos \theta + r \dot{\omega} $$
with the combined acceleration
$$ a = \sqrt{a_x^2+a_y^2} = \sqrt{ (g \sin \theta - r \omega^2)^2 + (g \cos \theta + r \dot{\omega} )^2 } $$
A slight simplification exists when constant speed is assumed ($\dot \omega = 0$)
$$ a = \sqrt{g^2 + r^2 \omega^4 - 2 g r \omega^2 \sin \theta} $$
