Is torque necessary in rigid body mechanics? Suppose we have a weightless, rigid rod fixed at one end, but free to swing at the other, where there is a mass $m$ attached.

If we want to determine the tangential acceleration of the mass using linear mechanics, we can use the formula,
$F = ma$
Thus,
$a = \frac{F}{m}$
Thus provided the force is always acting perpendicular to R, the mass will accelerate tangentially at $a$.

Now if instead we analyse the problem using torque, we can arrive at the equation,
$\frac{dL}{dt} = \tau =R\times F = RF$
Therefore, $\Delta L = RFt$ where t is the time the force is acting on the mass (again assuming the force is always applied in a direction perpendicular to R).
Since $L = mRv$, where $v$ is the tangential velocity of the mass, we have,
$mRv = RFt$
Therefore,
$\frac{v}{t} = \frac{F}{m}$
$a = \frac{F}{m}$
which is exactly as predicted from the linear analysis.

So in this very simplistic scenario, the acceleration of the body could be derived using either linear mechanics or utilising the concepts of torque and angular momentum.  Given this, in general when does the application of linear momentum start to break down when describing rigid body physics problems, and the concepts of torque and angular momentum become necessary to arrive at a solution?
 A: In your calculation, torque and angular momentum are not particularly advantageous because the example is too simple - the force acts in the direction of the velocity. If we had gravity force, the concept of angular momentum and torque would be already quite useful, as it allows us to write down the equation of motion of the rotary motion without the need to deal with the constraint forces from the rod acting on the particle.
In principle, torque and angular momentum are not necessary concepts in mechanics to formulate the laws of mechanics or equations of motion of a body. They are derived from the 2nd Newton law involving force and acceleration of material points.
Of course, it would be possible to express the state of motion of, say, satellite under action of an external force, with a set of coordinates and velocities of all particles that it consists of and write down the equations of motion for them involving only forces. But this would involve immense number of mutually constrained variables and would be very clumsy.
Introducing torque for each of the elementary force acting on every material particle of the rigid body, it is possible (under condition that force of one particle acting on another is directed along the straight line joining them) to simplify that large system of equations into only a few.
The state of the satellite can be described with 3 coordinates of its centre, 3 components of velocity of the centre, 3 angles of orientation and 3 components of angular velocity of the body, which is only 12 variables.
The easiest way to remember and write down the set of equations involving the angles of orientation of the body and the components of the angular velocity is the relation between net torque and angular momentum:
$$
\text{rate of change of angular momentum} = \text{sum of torques of external forces}.
$$
This can be used to derive the so-called Euler equations for motion of rigid body or some equivalent set of equations.
https://en.wikipedia.org/wiki/Euler%27s_equations_(rigid_body_dynamics)
The great advantage of the concepts of torque and angular momentum in practice is easier, more concise and more succint description of state of rigid bodies and of change of this state. It simplifies analysis of the mechanical problems.
A: You have placed some strong restrictions in order to develop your question. If you remove those restrictions, there will be many situations in which torque calculations are necessary. 
Consider a static equilibrium situation with a uniform board of length $\ell$, resting on two support points in a gravitational field. One support is at the left end of the board and the other support is $\ell /3$ from the right end of the board. To find the lifting forces of each support, you must you a torque calculation.
For another, moving situation, consider a force acting in a constant direction on a massive rod so that it's not constantly perpendicular to the rod, and the rod is not anchored. If the force isn't directed through the center of mass, the rod will begin to rotate, and the instantaneous tangential velocity of the end of the rod will not be $at$. You will have to calculate the torque on the rod to find $\omega (t)$. 
The precession of a spinning top in a gravitational field requires a torque calculation.  The explanation of steering/counter-steering of a bicycle or motorcycle driving down the road requires a torque calculation.
A: Torque is needed just as linear velocity is needed. Both indicate something happening at a distance. Torque is force at a distance, and velocity is rotation at a distance. In general, when you have one you also have the other.


*

*If you have a force through some axis, there is going to be a torque generated from that force away from this axis.

*If you have a rotation about some axis, there is going to be tangential velocity generated from this rotation away from this axis.

*A theoretical pure torque is a zero force applied at infinity just as a pure linear velocity is a zero rotation at infinity.


So in general we need torques because they tell us where things happen. A force applied along an arbitrary axis needs to be transferred to the center of mass as force/torque pair in order to be used in the equations of motion. Quite similarly any defined motion needs to be transferred to the center of mass as a linear/angular pair in order to establish the momentum state of the body (also for use in the equations of motion).
Consider the general case of your problem. A pinned rigid body at B is acted up a short duration force (impact) $F$ at point A. The center of mass is located at C. There is going to be a reaction at the pin $B_y$.

The equations of motion can be expressed at either A, B or C if we take care to account for the proper torques and accelerations.


*

*At the center of mass C the term $c \alpha$ is the linear acceleration of the center of mass and $(\ell -c) F$ the torque due to the force $F$.
$$ \begin{aligned} F + B_y & = m c \alpha \\
  (\ell-c) F - c B_y & = I_C \alpha \end{aligned} $$

*At the pivot point B the mass moment of inertia $I_C$ is transferred to the pivot using the parallel axis theorem
$$ \begin{aligned} F + B_y & = m c \alpha \\
  \ell F & = (I_C + m c^2) \alpha \end{aligned} $$

*At the impact point A there is an inertial load $m \ell \alpha$ and it's torque to account for (see last term of 2nd equation).
$$ \begin{aligned} F + B_y & = m c \alpha \\
  -l B_y & = (I_C + m d^2) \alpha - m d \ell \alpha \end{aligned} $$ where $d=\ell-c$


For all of the cases above the equations couldn't be solved properly without the equivalent torques on the left hand side and the equivalent accelerations on the right hand size. There are coordinate system locations where linear accelerations are zero (like the pin location), and others where net torque is zero (and therefore not needed). But not at the same time. You either have to account for one or the other or both.
A: It's very easy: you need to consider torques if you cannon treat your objects as mass points.
Here is a simplest situation:
Imagine a body and two forces working on it, the forces being antiparallel and of equal strength, but not on the same line.
The transaltional laws only tell you, that the center of mass will not accelerate, they cannot predict the angular acceleration. To ensure that the situation is static, you need also to check that all torques are zero.
It's the other way round: setting the torques with respect to enough axes to be zero implies also the net force to be zero.
In a two dimensional statics problem for example, you can either consider (i.e. set to zero) the net force and the torque around one point, or the net force projection on one direction and the torques around two points, or just the torques around three points. Each possibility gives you the complete information you can get out of the fact, that the situation is static (in the sense, that every additional equation will be linearly dependent of the existing ones).
