0
$\begingroup$

A famous kindergarten school experiment "The snatching cardboard and friction" intrigued me to ask this question.

Consider an empty glass placed on a table. Put a plane cardboard over the glass and put a coin on the cardboard.Now snatch the cardboard rapidly,you will find that the coin will fall downward in straight line freely under gravity till it has reached the bottom of the glass.

I think that when we applied the external force on the cardboard, the friction(either static or kinetic) between the cardboard and the coin hadn't originated because if friction had originated the motion of the coin would have been either (i) projectile motion or (ii) horizontal motion (along with the cardboard).

I think that friction is not simultaneous with external force. Am I correct or totally wrong?

$\endgroup$
2
  • 1
    $\begingroup$ I tried it, The coin I use falls down, but acquires rotation and always ends closer to the side of the pulled cardboard, sowing that some momentum is transfered by friction., so do you have a link? $\endgroup$
    – anna v
    Oct 1, 2019 at 11:38
  • $\begingroup$ @anna v I don't have the link right now but as you marked that the coin always ends closer to the side of pulled cardboard.It may be due to some other unspecified forces too as we don't know what is reality! $\endgroup$ Oct 1, 2019 at 15:59

1 Answer 1

0
$\begingroup$

Friction is present in this experiment, and it does operate while the card and the coin remain in contact.

The coin does not move very far in the direction the card moves in because contact lasts for only a very short time, so the friction force acts for only a short time. The sideways speed gained by the coin depends on the impulse it receives, ie the product of force and time of contact, rather than the product of force and distance moved relative to the card. So the quicker the card is pulled away, the less is the sideways speed acquired by the coin.

Another factor in the trick is that the card is quite thin so that the coin starts say only $0.254mm$ above the rim of the glass (this is the typical thickness of card). If the coin moves sideways while falling (like a projectile fired horizontally) then it only takes a fraction of a second for it to fall $0.254mm$ when it will be below the rim and will fail to escape from the glass.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.