Tipping while moving forward? (This is a follow-up to this question I asked earlier, but here let us consider a more general situation where the center of gravity is at a height $H$ above the base, and the force is applied at a height $h$.)

I understand that if the force applied is greater than $\dfrac{mgr}{h}$, i.e., if the force creates a greater torque than the weight force, the cylinder will tip.
If the force is lesser than that, but still greater than the maximum achievable friction $\mu mg$, it will slide.
My question is, what will happen if the force is greater than both $\dfrac{mgr}{h}$ and $\mu mg$?
Will it keep moving while tipping forward? Will it move some distance, and then topple down? Will it topple down first and then move forward?
How can we find the distance it would travel forward before toppling down entirely?
 A: If $F\gt f$ then there is a resultant force $F-f$ acting on the cylinder so its COM has a linear acceleration $a=(F-f)/m$ to the right. There is also a resultant couple, which is initially $\tau=hf-mgr$, acting clockwise about B. Hence the cylinder also has an initial angular acceleration of $\alpha=\tau/I$ where $I$ is the moment of inertia of the cylinder about B (not about its centre of mass E).
So yes the cylinder topples while also accelerating to the right. How long it takes the cylinder to topple, and how far it moves before this happens, requires a very tricky calculation.
If $\alpha$ were constant then the time to topple onto its side through angle $\theta=\frac12 \pi$ radians would be $t=\alpha/\theta=2\alpha/\pi$. And the distance moved by the centre of mass E would be $s=\frac12 a t^2$.
However there are complications. 
If $F$ is applied horizontally at a fixed point P then as the cylinder topples the vertical distance between P and B increases, while the horizontal distance between E and B decreases. So the torque $\tau$ and acceleration $\alpha$ do not remain constant; initially both increase as the angle which QE makes with the vertical increases. The increase in torque $\tau$ reduces the time $t$ which it takes for the cylinder to topple.
As the cylinder rotates its centre of mass E accelerates upwards. To enable this to happen the normal force $N$ acting at B must also increase. This in turn increases the friction force $f$ which in turn increases the torque on the cylinder, causing it to topple faster, and decreases the horizontal acceleration of E. 
The simplified calculation above also assumes that point B remains in contact with the ground as the cylinder topples. But it is possible for the rotation of the cylinder to increase so much as it topples that at some stage point B leaves contact with the ground. Then $f$ becomes zero and the torque on the cylinder changes from clockwise to anticlockwise. It continues rotating, but now about E.
These factors affect each other, and if the cylinder leaves contact with the ground then its centre of rotation changes and it becomes a projectile until it regains contact with the ground. So it is a very tricky calculation.
