Expression for angular friction Consider say a door rotating about its axis. Is there, in general, any expression for the frictional hindrance to its motion? I was thinking in line with the coefficient of friction for linear motion on a surface.
 A: Dry Friction works in the same way for rotational motion as it does for linear motion, except that it now causes a torque which opposes relative rotation. There is still static and kinetic friction which depends on the normal force $N$ between the surfaces in contact, according to the same empirical law $F \le \mu N$. 
There are several ways in which friction can arise in rotation, depending on how contact is made between the moving parts. Some examples :


*

*A door is hanging vertically on a loose horizontal hinge, as in Calculating the time to stop a wheel with friction. Here the area of contact is at a fixed distance $r$ from the axis of rotation. We don't need to calculate this area, we don't need to consider that the normal force varies across the contact area. The friction force depends only on the total normal force $N$. The torque due to kinetic friction is $\mu Nr$. If the door hangs at rest then $N=Mg$ is the weight of the door. 

*Similar is a door resting on a vertical hinge passing symmetrically through its centre of mass. The weight of the door is again supported by vertical normal forces at (approximately) a fixed distance from the axis.

*The door hangs from a vertical hinge on one side. In addition to vertical contact forces to support the weight of the door (as in #2), there are also horizontal contact forces providing a torque to prevent the door from rotating about a horizontal axis. The total frictional torque is greater.

*A tight hinge presses radially inward on the axle. The friction torque is again $\mu Nr$ but now $N=2\pi T$ where $T$ is the 'hoop stress' in the hinge.   
In each of the above cases, if the door is rotating or oscillating then the normal contact force increases to provide centripetal acceleration.


*A horizontal disk rotates about its centre on a rough horizontal surface. The normal force is spread uniformly across the disk. However, friction near the rim exerts more torque than near to the axis of rotation, so the torque must be integrated with respect to radius. The result is a torque of $\frac23 \mu Nr$. See Rotational physics of a playing card. 

A: Friction only appears to stop the motion. It will not point radially but rather backwards on the sliding area. That is, it will point opposite to the door swinging direction and not towards the door hinge. Because nothing is moving/sliding in the radial direction. So, the concept of an "angular friction" is nothing more than usual "linear" friction. The only difference is that this usual "linear" friction then causes a torque that counter-acts the swinging - but neither the kinetic friction formula $f_k=\mu_k n$ nor the torque formula $\vec \tau=\vec f_k \times \vec r$ are any new particular "angular" friction expressions.
Since this usual "linear" kinetic friction $f_k=\mu_k n$ does not depend on sliding speed and also not on contact area, then for a rigid door, it will be a constant force regardless of swinging speed or size of hinge. This force will then be applied at some averaged position. If the whole bottom of the door slides, then applied half-way; if only the hinge slides, then applied half-way across the hinge.
So, friction itself does not change, but if you can concentrate it near the rotation point (the hinge), then the counter-torque caused by this friction will be small and not have a large influence.
A: Not in terms of dry (coulomb) friction. But in terms of damping, relating angular speed to torque $$\tau = c_\theta \,\omega$$
where $c_\theta : [\rm \frac{N m}{rad/s}]$
The source of this damping is a thin lubrication layer and the corresponding shearing of the lubricant. It is easy to estimate the damping coefficient from the geometry and the viscosity of the fluid.
