kinematics - Rotational physics of a playing card

A playing card leaves a dealers hand with some angular velocity. As the card slides across a table the friction of the table causes the rotation to slow. How is the friction coefficient between the card and the table related to the angular acceleration of the card? IE how can I calculate angular acceleration?

Answer

I think this is a good question, but difficult to answer.

Frictional coupling between sliding and spinning motion includes a mathematical treatment of this situation. The solution involves elliptic integrals. The rotating and translating motions are coupled through friction, so that the disk always stops rotating and translating at the same time. The same thing happens with playing cards - the phenomenon is not peculiar to disks. There is a more accessible description in the Focus section of the American Physical Society website.

Calculating the effect of friction on a disk which is translating but not rotating, or a disk which is rotating but not translating, are both relatively straight-forward. But the combination is much more difficult to analyze.

For example, a disk of radius $R$ and mass $M$ translates on a flat horizontal surface with dry friction $\mu$. The friction force is $F=\mu Mg$. The linear deceleration is $\frac{F}{M}=\mu g$.

If the disk is rotating without translating, the friction force creates a torque. $F$ is distributed evenly across the disk but the torque it exerts depends on distance $r$ from the axis of rotation. The area of an annulus of thickness $\delta r$ at radius $r$ is $2\pi r \delta r$, the friction force on it is $\frac{2\pi r \delta r}{\pi R^2}F=\frac{2r\delta r}{R^2}F$ and the torque is $\frac{2r^2\delta r}{R^2}F$. Integrating from $r=0$ to $R$, the torque on the whole disk is $\tau=\frac23RF$. The moment of inertia of the disc is $I=\frac12MR^2$ so the angular deceleration is
$\alpha=\frac{\tau}{I}=\frac{\frac23R\mu Mg}{\frac12MR^2}=\frac{4\mu g}{3R}$.