One Rupee Mystery: Why Does the Coin Fall and Card Move?

[ahsen_tahirpk]

Put a one rupee coin over a piece of card paper placed on an empty glass push the card with a sudden stroke of finger card will move ahead while the stone fall in glass, Why?

 

[Rap]

Because the coin slides on the card. If you put the card on the table and push the coin, it will slide on the card the same way. If you glued the coin to the card, and pushed the coin, both the card and the coin would move together. If you glue the coin and the card together and put them on the glass, and push the card, both will move off the glass.

When you snap the card, it doesn't push on the coin very much, so the coin stays there, and when the card is gone, the coin drops.

 

[ashishsinghal]

but you have to move the card really quickly so that there is very less time for friction to impart velocity to the coin.

 

[m.e.t.a.]

This trick relies on the fact that the coefficient of kinetic friction, \mu_{k} is not proportional to sliding speed v. If you double the sliding speed, the resulting coefficient of kinetic friction less than doubles.

To understand why this is important to the card slide trick, let's see what would happen if \mu_{k} were proportional v:

Scenario 1
You pull the card out from under the coin at some constant speed v_{1}. The card and the coin are in contact for time period T_{1}; the frictional force is \mu_{k,1}. The total impulse imparted to the coin due to friction is then I_{1}=mg \mu_{k,1} T_{1}.

Scenario 2
You pull the card out at speed v_{2}=2 v_{1}. The card and the coin are now in contact for half the time (T_{2}= {\textstyle{1 \over 2}} T_{1}) but with double the frictional force (since we are assuming \mu_{k} \propto v): \mu_{k,2}=2 \mu_{k,1}. The total impulse imparted to the coin due to friction is then I_{2}=mg \mu_{k,2} T_{2}=mg (2 \mu_{k,1}) ({\textstyle{1 \over 2}} T_{1}) = I_{1}.You can see that, if \mu_{k} \propto v, the coin would always be given the same impulse no matter how fast you pulled the card out. In the real world, however, \mu_{k} and v are not proportional to one another, but are related approximately linearly by \mu_{k} = Av+\mu_{k,0}, where A and \mu_{k,0} are positive constants. Therefore, in the real world, the faster you pull the card out, the less impulse is imparted to the coin.

I found some experimental data http://idol.union.edu/vineyarm/teaching/phy110lab/sample_report_2.pdf" which show how \mu_{k} varies with v for different materials.

 

[rcgldr]

This trick relies on the fact that the coefficient of kinetic friction, \mu_{k} is not proportional to sliding speed v. If you double the sliding speed, the resulting coefficient of kinetic friction less than doubles.

In most cases, kinetic friction decreases slightly with sliding speed, even in a vacuum where there's no effect from air.

The other issue is that the maximum rate of acceleration of the rupee is the kinetic coefficient of friction between it and the card. For example, maybe it's 1/2 g (1/2 of 9.8 m/s²). If the card is accelerated at 4g's or more, then the rupee's acceleration is still 1/2 g (or a bit less), and the card passes by the glass well before the rupee does, so the rupee falls into the glass.

The key factor is if the card accelerates and/or moves fast enough so that it clears the glass before the rupee has enough time to clear the glass due to it's limited acceleration related to kinetic coefficient of friction (probably less than 1/2).