Do we take torque/angular momentum about a point or an axis? It has left me a bit confused because at some places the angular momentum/torques are taken about a point, while in others, it's taken about an axis.
Consider the 2 situations:
Situation-1: In a conical pendulum rotating with constant $\omega$, taking the angular momentum about the point from where it has been attached to the roof, yields that the magnitude of the angular momentum is constant, but direction keeps changing.
Situation-2: Here forces were acting on the endpoints of the rectangle and for it to be in equilibrium, Torque was balanced about the diagonal. (The solution specifically mentioned balancing the torque about the diagonal as axis)[The question was about the suspension of cars when one of the tires is raised a bit
higher on the pavement.]
 A: Well consider the equation of moment of momentum (angular momentum) and moment of force (torque)
$$ \begin{aligned} \vec{L} & = \vec{r} \times \vec{p} \\ \vec{\tau} &= \vec{r} \times \vec{F} \end{aligned} $$
But what is $\vec{r}$? It is the location of the percussion axis or line of action relative to the reference point.
Momentum and forces act along a line in space, that when offset from a reference point, cause the moment of momentum or moment of force. The trick is that the reference point is a point in space, but forces and momentum act anywhere along their lines. 
This means that if you take any $\vec{r}$ in the direction of $\vec{p}$ or $\vec{F}$ the result is the same (since the cross product removes all parallel components).
The same applies to velocities which are resolved at a reference point, and they are the moment of rotation with $\vec{v} = \vec{r} \times \vec{\omega}$. The motion is a line in space called the rotation axis.
In Summary
In mechanics, there are three lines in space, one for motion, one for momentum and one for forces, and their moments ($\vec{r} \times \vec{\rm line}$) are resolved about a point.
Some of the common points of reference are


*

*The center of mass.

*The origin point.

*A kinematic joint location.

A: Moment of a force about a point is the concept usually used by physicists. Moment of a force about an axis is, I believe, a different concept, used more by engineers than physicists.

Consider the moment of a force, $\vec F$, acting into the page, with a line of action through point C. Its moment about point P is
$$\vec {\tau_P} = \vec r \times \vec F= \text {a vector of magnitude}\ rF\ \text{along DB.}$$
Its moment about point O is
$$\vec {\tau_O} = \vec s \times \vec F= \text {a vector of magnitude}\ sF\ \text{in the direction shown.}$$
What is the moment of $\vec F$ about the axis BD? It is surely $\vec {\tau_P}$, in which P is chosen so that the displacement $\vec {PC}=\vec r$ is perpendicular to the axis (as well as to $\vec F$).
We can also obtain the moment of $\vec F$ about BD by taking the moment about an arbitrary point on the axis, such as O, and then taking the component of this moment along BD.
Magnitude of this component = $sF \cos \theta = rF$.
In summary...
If the direction of the axis is represented by the unit vector, $\hat a$, and $\vec s$ is the displacement vector from a point on the axis to a point on the line of action of the force, $\vec F$, then
$$\text{Torque of $\vec F$ about axis}=\hat a\cdot (\vec s \times \vec F)=\vec F \cdot (\hat a \times \vec s).$$
A: In general, torque and angular momentum are taken about a point.  For the special case of rotation of a rigid body about a fixed axis (or more correctly rotation in a plane if the axis moves, such a ball rolling down an incline), torque and angular momentum are taken about the fixed axis.
A: Following the definitions of torque and angular momentum $$\vec{L}\equiv \vec{r}\times \vec{p}$$ $$\vec{\tau}\equiv \vec{r}\times \vec{F}$$ both are evaluated about a point: the origin, from which you take your position vector $\vec{r}$.
Now in Situation-1 if you take your origin at the pivot of the pendulum then indeed your angular momentum lies in the plane of the string being perpendicular to it, and rotates together with it. This dynamics is consistent with the equation of motion $$\frac{d \vec{L}}{dt}=\vec{\tau}$$ if you compute the torque counting $\vec{r}$ from the same origin (the pivot; first rule of calculation: be consistent and stick to the convention throughout the calculation), you will see that torque is again always perpendicular to the string but lies in the plane normal to the direction of Earth's gravity. $\vec{L}$ and $\vec{\tau}$ are therefore always orthogonal. Hence $\vec{L}$ is perpetually precessing while staying constant in magnitude.
If you choose your origin in the center of the circular trajectory of the mass. The picture will be superficially different. The angular momentum will now be constant and look perpendicularly up. The torque, - as computed from the new origin, - will be zero (in the steady state motion the net force on the mass points towards the center of the circular trajectory), in consistence with the equation of motion above.
As for the Situation-2, this in my opinion, is more of an Engineer's thing (as the context suggests). Repeating again that, strictly speaking, torque can only be defined with respect to the origin (a point), it is often convenient to talk of the torque about an axis $\vec{h}$, meaning a projection of the torque $\vec{\tau}$ on this axis $\tau_{axis}\equiv \tau \cdot \vec{h}$. This is handy as in practical applications you most often encounter situations when you are only interested in motion around a fixed axis. In this case indeed only the projection of total torque along the axis matters, since all radial forces are supposed to be cancelled out by the reaction from the rod defining the rotation axis (aka the shaft).
However one should cautious when using this notion of $tau_{axis}$ as in situations where no rigid axes are present (like in the conical pendulum case), then radial forces are not cancelled out, and $\tau_{axis}$ is ill-defined (depends on where you choose your origin).
