If we take the free body diagram above and sum the moments about the
center of mass, we would find that an increased applied force would in
fact cause the solid body to rotate.

Perhaps. Or additional forces can appear. If I push up on my car's bumper, a rotational force is being applied. But the normal force on the wheel farther from me increases, so the total torque still sums to zero.

The problem with this understanding is that during rotation, the
center of mass of the drawn system should actually rises relative to
the surface the bike is moving across. If the bike were to be truly
rotating about its center of mass, then the back wheel would begin to
dip below the surface of the road like you might see in a glitchy
video game.

Another way to interpret this is that if you apply a small torque and imagine it around the center of mass, you're pushing the rear wheel into the ground. As you do so, the normal force increases. This increased normal force counters the torque you are applying. But the maximum this can be is the weight of the bike. So if you increase past this maximum, the bike will rotate.

I had originally tried to analyze this from the rear axle, but because the bike will accelerate, this makes fictitious forces appear in the axle's frame that have to be dealt with. We can mostly ignore this by analyzing around the center of mass instead.

I'll ignore friction for now, and just assume that we have sufficient friction to avoid wheel slip. Then the forces we need to consider are the weight, the normal forces, and the frictional force from the road.

As the wheel accelerates faster it provides a torque to the bike. The bike responds by changing the balance of the normal forces. At the limit, only the rear wheel is providing a normal force, and that will equal the weight of the bike. Since gravity acts through the center of mass, the only torques that appear are the normal force and frictional force. When torque from friction exceeds torque from the normal, the bike will tip.

$$\tau_{friction} > \tau_{Normal}$$
$$F_f \times y > F_N \times x$$
$$F_f > \frac{mgx}{y}$$

To tip the bike, the wheel has to push with a force greater than the weight of the vehicle times a factor that depends on the location of the center of mass. \

And since we have the forward force, we can solve for forward acceleration and know that as it begins to tip, the bike will be accelerating at $\frac{x}{y}g$

And then the bit that I think began your question:

...the only forces acting about the rear axle, point O, is the force of weight. This means that an increase in applied torque and subsequently, the
applied force Fa, should not effect the rotation about the rear axle.

In your initial diagram, you were neglecting one additional force, and that is the fictitious force due to acceleration of the frame. This force is equal to $ma$ and is applied to the center of mass in the direction opposite the acceleration of the bike. As the acceleration is due to this force, it means it does affect the rotation, even though it wasn't obvious when you started summing torques.

We refer to the figure above. Provided the rod is firmly fixed at ‘A’ and ‘B’, i.e. there is no sliding at ‘A’ and ‘B’, although it may swivel as on hinges, and gravity acts downwards in the figure (effective as force $F_C$), the supporting forces at ‘A’ and ‘B’ will always be $F_C/2$. Changes in θ do not change the relationship. Even in the vertical state when ‘A’, ‘B’ and ‘C’ lie on a vertical line, the rod is equally supported at ‘A’ and ‘B’. We apply some statics.

Setting A as the centre of rotation, the vertical forces are:
$$F_BL\cos\theta = 0.5F_CL\cos\theta$$
$$therefore\qquad F_B = 0.5F_C$$

Similarly, setting B as the centre of rotation, the vertical forces will are:
$$F_AL\cos\theta = 0.5F_CL\cos\theta$$
$$F_A = 0.5F_C$$

Now if, for example the fix point at A is a slot which allows the rod to not only swivel, but also to move freely in the longitudinal direction, then:

- Nothing changes at $\theta$ = $0^{\circ}$ since the rod will experience the
full normal (upward) force at ‘A’.
- At $\theta$ = $90^{\circ}$ however, Fix point ‘A’ would merely keep the rod
from rotating in case of a slight imbalance in sideways forces on the
rod, which is common in the real world, and point ‘B’ would bear the
full weight of the rod.
- At $\theta$ between $0^{\circ}$ and $90^{\circ}$, provided friction at 'A' was
not very high, 'B' would bear more than half the weight of the rod,
since point 'A' would merely be required to prevent the rod from
rotating left and falling towards the horizontal. The vertical
component borne at 'A' would be less. As $\theta$ increased, 'B'
would bear an increasingly greater burden, until it bore all the
weight at $\theta$ = $90^{\circ}$. The ratio of the weight it bore
would be proportional to the friction coefficient of the slot at 'A'.
A high enough frictional coefficient, like the feet of those lizards
that walk on walls, at 'A' would cause 'A' to bear half the weight
even at $90^{\circ}$, just as in the original condition above.

## Best Answer

There is an old trick that turns a dynamics problem into a statics problem. Apply equal an opposite inertial forces on the center of mass.

Lets look at a

free body diagramof a motorcycle accelerating with $\ddot{x}$.The balance of forces in the horizontal direction equate the tractive force $B_x$ to the acceleration $$ \left. m \ddot{x} - B_x = 0 \right\} B_x = m \ddot{x} $$

The balance of forces in vertical direction, together with the balance of moment about any point (I choose the rear contact point) give us the distribution of loads on the tires.

$$\left. \begin{align} A_y + B_y - m g & = 0 \\ -\ell A_y + c m g - h m \ddot{x} & = 0 \end{align} \right\} \begin{aligned} A_y &= \frac{c}{\ell} m g - \frac{h}{\ell} m \ddot{x} \\ B_y & = \left(1-\frac{c}{\ell}\right) m g + \frac{h}{\ell} m \ddot{x} \end{aligned} $$

ALSO, if you want to find the minimum acceleration for a wheelie then set $A_y=0$ to find $$\ddot{x} \ge \frac{c}{h} g$$

FINALLY, if traction is limited to $B_x \le \mu B_y$ then the equations above give us the maximum acceleration before traction is lost $$\ddot{x} \le \mu g \frac{\ell-c}{\ell-\mu h} $$