Answer to Question #89788 in Mechanics | Relativity for David D. Duobah

Question

A popular magic trick consists of placing a coin on a card and the card on the top of a glass. The edge of the card is flipped briskly with the forefinger, causing the card to fly off the top of the glass as the coin drops into the glass. Explain. What law does this illustrate?

Answer 1

Consider there is friction between the card and the coin. The force moving the coin with the card is the force of static friction between the coin and the card.

According to Newton's second law in terms of momentum, this force (acting for $\Delta t$ seconds, time of your finger's contact with the card) is:

$$F_{sf}=\frac{m\Delta v}{\Delta t},$$

$$\Delta v=\frac{F_{sf}\Delta t}{m},$$

where $m$ - mass of the coin and $\Delta v$ - change in the coin's speed (from 0 to some speed).

The distance that the coin can travel in these conditions (is we assume constant acceleration) is

$$d=\frac{1}{2}\Delta v\cdot\Delta t=\frac{F_{sf}\Delta t^2}{2m}.$$

Expand the force of friction between the coin and the card:

$$d=\frac{\mu mg\Delta t^2}{2m}=\frac{\mu g\Delta t^2}{2}.$$

As we see from the last equation, because of the brisk flip the time is so short that the force of friction accelerates the coin to very small speed and for the time of $F_{sf}$'s action the coin can't cover the distance from the center of the glass to its edge. However, if you take a very narrow glass, or very shaggy card or flip the card not so briskly, the trick can fail.