Forces on wheels in an accelerating vehicle Assume that a motorcycle of mass $m$ has two wheels that are equidistant from its centre i.e the force on each wheel is $mg/2$.
If the motorcycle accelerates forward, will the two forces on each wheel (measured instantaneously) remain the same? If not, how can one mathematically describe the change in forces measured on each wheel and will these forces oscillate before converging when the motorcycle reaches final velocity?
Now, if the weight of the motorcycle isn't uniformly distributed along it's length e.g force measured at one wheel is say $mg/3$ and $2mg/3$ at the other, how different would the vehicle dynamics (oscillations etc) be? Can one somehow infer the fact that the weight isn't uniformly distributed without actually measuring the forces at the wheels?
I know this question is very vague but any ideas are very welcome. 
 A: First of all geometric center and the center of mass are two different things.
Since a motorbike would not have center of mass in the very middle due to the facts that each part of the bike weighs different as well as the riders position. You would need to do a free body diagram to start your analysis. Then you can analyze the moments around the points of contact of each wheel then get the correct force value. If it was uniform object with wheels with equal distance to center of mass, your mg/2 assumption would be correct, however, it would be best to analyze the moments around the point of contact to find correct reaction forces.
Also, as the bike starts to accelerate, there will be forces applied through the front or rear wheel. This would change the balance of forces on the wheels as there are more forces present.
My advice to you is to, draw a FBD for each scenarios and then analyze the reaction forces. Through finding reaction forces, you can find shear and normal stresses which would help you to find oscilations. However this is not as simple forward as you put it in the question.
A: 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 diagram of 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 $$\ddot{x} \le \mu g \frac{\ell-c}{\ell-\mu h} $$
