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Here is a simple schematic of the engine.
enter image description here

The increase in angular momentum of the crankshaft in one complete cycle would be

$\Delta L_{cycle}=\int_{\delta =0}^{\delta =4 \pi} F cos \theta Rdt$.

where F is positive in combustion stroke and negative in other strokes (as L increases only in the combustion stroke and decreases in other strokes where work is done by the piston to draw in air-fuel mixture, compress the mixture, or expel exhaust gas). Also note, F would be different for each value of 𝛿 from 0 to 4𝜋.

The torque generated by the engine would then be $$\tau = \frac{\Delta L_{cycle}}{\Delta t_{cycle}}$$

Now, let's see what happens when the angular velocity (ω) of the crankshaft is doubled.
Any force F specific for an angle 𝛿 would then act on the crankshaft for only half the period of time (i.e., dt for each angle 𝛿 gets halved). As a result, the increment in angular momentum of the crankshaft in one complete cycle gets halved. However, since the time period of each cycle has halved as well, the torque generated by the engine should theoretically be the same regardless of the rpm of the engine. This would result in the following torque-rpm graph.

enter image description here

However, we know that this is not the case. The actual graph looks something like this. enter image description here The dip in torque is easy to explain as with increasing rpm one can expect several losses to increase such as frictional losses, pumping losses, inefficient charge mixing, decreased time for efficient combustion etc. But my question is, what explains the increase in torque with angular velocity at the low rpm range?

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Your assumption that the force acting on the crankshaft for a given crank angle $δ$ is constant regardless of the rpm of the engine is false.

A primary reason why this is the case is, unlike what we generally assume, the intake and exhaust valves don't open or close exactly at TDC (top dead center) or BDC (bottom dead center). The timing of the opening and closing of the two valves is tuned according to the requirements of an engine as @Niels said earlier. Consequently, an engine will produce the maximum torque only at a certain rpm.

Here is a graph depicting the above idea. enter image description here

Out of the four valve timings, the intake valve closing time (IVCT) is the one that heavily influences the torque-rpm curve of an engine. This is because the torque-rpm curve closely follows the volumetric efficiency (VE)-rpm curve, which is dictated by the IVCT.

enter image description here

VE of an engine is the ratio of the mass of fresh charge trapped by the cylinder to the mass of fresh charge that would occupy the swept volume of the engine at atmospheric pressure.

Charge intake occurs even after BDC of the intake stroke and continues till the early compression stroke. This occurs because of the negative difference in between the pressure of the cylinder and that of the intake manifold at BDC (the difference increases in magnitude with increasing rpm) as well as the inertia of air flow in the intake system (causing air to be drawn in even after the cylinder pressure equates with that of the intake manifold).
The maximal mass of charge is trapped when the crankshaft has rotated a few degrees after BDC. If the intake valve remains open even after this point, the charge in the cylinder starts to get expelled from the intake valve, which is obviously disadvantageous. This optimal timing of the closure of the intake valve varies with the rpm of the engine (for higher rpm, the optimal timing arrives more and more late after BDC).

Conventional engines don't have variable valve timing, i.e., like other valve timings their IVCT is fixed as well. The engine's IVCT is consequently optimal for only a single value of rpm––the optimal rpm.

When the rpm of the engine exceeds this optimal rpm, the mass of charge drawn in during intake becomes lesser and lesser. Whereas, when the rpm is below the optimal rpm, the charge starts getting expelled before the intake valve closes. Thus, an engine running in any rpm other than its optimal rpm would have a lower volumetric efficiency. And if you combust a lesser mass of charge, you are surely to generate a lesser amount of torque.

Here is a good read on valve timings and their impact on engine performance.
The impact of valve events upon engine performance and emissions.

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This is because the valve timing is optimized for a certain RPM, which represents the design point of the engine. In the case of a piston engine for use in an airplane, the opening and closing of the intake and exhaust valves is timed for optimum performance (i.e., maximum work extracted from the combustion process) at 2400-2600 RPM, while on a Suzuki GS1000 motorcycle engine it is instead set to accomodate 10,000 to 12,000 RPM.

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In addition to the volumetric efficiency of the valve timing system, there are some harsh losses associated with thermal losses. The top of the cylinder is hotter than the bottom end, and heat transfers down the cylinder liner at a rate that doesn't depend on rpm. So more heat is wasted per stroke (which is a poor way of saying it) when the engine is running slowly.

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