Possible for Magnus effect to overcome rotational phenomena? I’ve noticed that when a moving, rotating disk is in contact with a surface, it curves the opposite way it is spinning (eg, a puck on a table moving forward and rotating clockwise would curve left). Interestingly, this is opposite to the Magnus effect.
As for the cause of this, the best I can figure is that as an object moves, the friction causes it to put more pressure on the front than the back as it tries to lean, which gives more ‘weight’ to the sideways friction in the front than the back, causing it to curve.
On to my real question: assuming this is the cause of the phenomenon, and since velocity does affect fluid friction but not solid friction, would it be possible to spin the object so fast the Magnus effect overcomes this phenomenon to make the disk curve the opposite direction?
 A: You are correct in the cause of the lateral drift of the cylinder, the friction force gives rise to a torque with respect to the centre of mass of the cylinder. This must be compensated by an increase in pressure at the front of the contact area and a decrease at the back, which in turn generates a force perpendicular to the direction of translation. If $\textbf{e}_v$ and $\textbf{e}_\omega$ are the unit vectors in the directions of the translational and rotational velocity, respectively, then the direction of the force is given by $\textbf{e}_v\times\textbf{e}_\omega$.
Let's consider a cylinder of radius $R$, height $H$ and mass $m$ sliding with a velocity $v$ whilst rotating with angular velocity $\omega$. We also define the velocity ratio
$$
\epsilon = \frac{v}{R\omega}.
$$
It can be shown [1] that the magnitude of the perpendicular friction force is given by
$$
F_p = \frac{\mu^2mgH}{R}f(\epsilon),
$$
where $\mu$ is the coefficient of friction and $f(\epsilon)$ is a complicated function involving elliptic integrals. All we need to know about $f(\epsilon)$ however is that $f(0)=0$, $\lim_{\epsilon \to \infty} f(\epsilon)=0$ and $f(\epsilon)$ is positive and finite for all $\epsilon>0$.
We can now consider the Magnus effect. The magnitude of the force induced by the Magnus effect is
$$
F_M = 2\pi\rho vHR^2\omega,
$$
where $\rho$ is the air density. In order for the force due to the Magnus effect to overcome the perpendicular friction force, we require that $F_M>F_p$ i.e.
$$
2\pi\rho vHR^2\omega>\frac{\mu^2mgH}{R}f(\epsilon)
$$
or
$$
\frac{2\pi\rho R^3 v\omega}{\mu^2 mg}>f(\epsilon).
$$
Given that for all $v,\omega>0$, $f(\epsilon)>0$ and finite, this means theoretically we can always ensure the inequality is met by changing the physical parameters, for example by making the cylinder have a large radius, or making it very lightweight.
For $\epsilon=1$, $f(1)=512/135\pi$ and we require that
$$
\frac{\rho R^4\omega^2}{\mu^2 mg}>\frac{256}{135\pi^3}\approx0.06.
$$
Conversely, whatever the physical parameters of the cylinder, air and surface, by making $\epsilon$ small enough you can always ensure the inequality is satisfied and the Magnus effect will outweigh the perpendicular friction force.
