What rotates a door, torque or force? I am confused as to what causes a door to rotate. Assume I apply a force to the edge of the door, with the door initially at rest. Force is perpendicular to the door, so it would create a torque. But isn't there also a force on the door? The hinge itself sees very little force, so isn't the door getting a force on the centre of mass and a torque? Why doesn't the door then rotate AND push out of the frame?
Take, for example, a sphere in space. If you apply a force, it will translate according to $F=ma$, but it will rotate according to $\tau=I\alpha$.
So for the door, doesn't the door have both a torque and force on it? Which is the cause of its rotation, and how does a net force on the door work then?

 A: There are three forces on the door: gravity $\vec F_g$, the applied force $\vec F_a$, and the constraining force from the door hinges $\vec F_h$.  $\vec F_g + \vec F_a + \vec F_h = m\vec a$ where $a$ is the acceleration of the center of mass (CM)  of the door and $m$ is the mass of the door.  The component of the hinge force in the vertical direction counters both the force of gravity and any component of the applied force downward, if there is any. The hinges provide a constraint causing rotational motion about a fixed axis vertical up through the hinges. Taking the torque about this axis $\vec \tau = \vec F_a \times \vec d = I \vec {\dot\omega}$ where $I$ is the moment of inertia of the door about the axis of rotation, $\vec \omega$ is the angular velocity of the door about the axis, and $\vec d$ is the vector from the axis of rotation to the point on the door where $\vec F_a$ is applied.  The hinge force does not have a torque about this axis of rotation since the moment arm for the hinge force is zero, and gravity does not provide a torque about this axis.  The CM motion is based on the net force, rotation of the door about the fixed axis of rotation is based on the net torque.  Both forces and torques contribute to the overall motion of the door.  The net force is greater than zero for the CM to move. The net torque about the axis of rotation is greater than zero for the door to rotate about the axis. Both
a net force and a net torque cause the motion.
I suggest you draw out a free body diagram showing the forces and torques; use polar coordinates about  the axis of rotation.  For simplicity, assume $\vec F_a$ has no vertical component and is always perpendicular (90 degrees) to $\vec d$.
A: Consider a door floating in outer space, not attached to anything.  If you apply a force to the door at its center of mass and perpendicular to the door, it will be displaced in the direction of the force with an acceleration given by Newton's 2nd Law, but it will not rotate.  If, on the other hand, you apply a force perpendicular to the door and at the edge of the door, it will rotate AND be displaced.
The difference between the above scenario and a door in your house is that the door in your house is on a hinge.  When you push on the edge of the door opposite the hinge, the door only rotates because the hinge is applying enough force to keep the door from translating.  Even though that force is "small", it still exists.
This all means that any force that is not through the center of mass of an object such as a door will cause a torque that will rotate that object unless there is another torque that stops that rotation from happening.  That is the "normal" case, and the case of a force causing only translation is a special case where the summation of all torques leads to zero net torque.
A: It is possible to analyze the total movement as a sum of rotation of the COM around the hinge, plus rotation of the door around the COM.
Considering the door initially at rest, the sum of the forces must be greater than zero to move it. $\mathbf F_a + \mathbf R_h = m\mathbf a_{com}$, where $\mathbf F_a$ is the applied force and $\mathbf R_h$ is the reaction at the hinge. The later can be taken as a sum of a component along the door (parallel) and another normal to the door.
$\mathbf R_h = \mathbf R_{h\perp} + \mathbf R_{h\parallel}$
Considering the applied force $\mathbf F_a$ perpendicular to the door, the torque around the COM is:
$\tau = r_h R_{h\perp} + r_a F_a$, where $r_h$ and $r_a$ are the distances from the COM to the hinge and $\mathbf F_a$ respectively.
As a result of this torque, the door acquires an angular acceleration around the COM:
$\tau = I\alpha$
But at the same time, due to the centripetal force $R_{h\parallel}$, the COM rotates around the hinge. $$\frac{R_{h\parallel}}{m} = \omega^2 r_h$$ where $m$ is the mass of the door and $\omega$ is the instantaneous angular velocity around the hinge.
Of course, as the hinge is a constraint of movement, the combination of rotation of the COM around the hinge, plus the rotation of the door around the COM results in a rotation of the door around the hinge.
