# 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?

• Why would the friction coefficient depend on the angular acceleration at all? – ACuriousMind Dec 18 '16 at 11:45
• The greater the friction coefficient the faster it's going to decelerate? – Vindictive Dec 18 '16 at 11:48

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
$\alpha=\frac{\tau}{I}=\frac{\frac23R\mu Mg}{\frac12MR^2}=\frac{4\mu g}{3R}$.

• New thought, if the card always stops rotating and translating at the same time. Can I not simply calculate the duration t of the translation using an initial velocity v and linear deceleration F/M = μg. And then with some initial angular velocity w use the time t to calculate the angular acceleration α= w/t? – Vindictive Dec 19 '16 at 17:03
• No. That is the point of the research. You cannot uncouple the motions. As the Focus article says : a spinning disk experiences less friction and slides farther than a disk without rotation. So your calculation will not give you an exact result. – sammy gerbil Dec 19 '16 at 22:04

When you see an object "slide" on another, it means there "kinetic friction" is playing role.

Kinetic friction, like other forces, is a force exerted on a certain part of the card. The special property of friction is that It opposes relative movement of objects. No matter you consider the force itself, or the torque that is exerted to the object, It will oppose relative movement of masses.

So (not exactly talking) the force acting on the center of mass of the card is in the direction opposite to its motion So the linear movement of card is slowed as it moves ahead.

The torque acts the same! In the direction opposite to that of angular velocity. So again it reduces the angular speed and slows the card rotation down.

These slowing downs continue until the card is full stopped. When there is not movement for friction force to oppose.

• Okay, I think your statement can be summed up as: friction opposes movement both linear and rotational. Fine, I know. I'm wondering how I can use the friction between the card and table to calculate the rate at which the angular velocity slows, IE the angular acceleration. Or how can I use the linear velocity to calculate the torque being applied to the card so that I can again calculate the rate at which the angular velocity of the card is changing. – Vindictive Dec 18 '16 at 11:57
• @Vindictive For those, I need a specification of the shape, mass and mass distribution of the card. And also the way it hits the table! I couldn't go further with these information. Add them. I'll edit. – AHB Dec 18 '16 at 11:59
• A playing card you can generalize as a rectangle with width w and height h. mass of m and mass distribution of equally distributed throughout so the centre of mass is at h/2 and w/2. You can ignore the the part where the card "hits" the table. – Vindictive Dec 18 '16 at 12:04
• @Vindictive To be honest, now that I am doing it on paper, I see I am incapable of solving the angular part. And I don't want to give you wrong information. So I leave it up to others. But I can explain the linear part (Center of Mass) of the problem if you want. – AHB Dec 18 '16 at 12:34