If you hit the lowest coin in pile of coins fast enough it does not fall due to inertia. Why does it fall if-?

Question

You hit it slowly. The inertia of the pile of coins is still there so only one should move? What has the speed of hitting got to do with inertia of rest?

Just to be more concise could anyone cite a mathematical relation that would relate crtical time need before which moment of inertia of rest wont act at all. That if mass coin of mass m moves with x m/s the upper coins havin mass n X m won't fall.

I would request A_Bank to explain change in momentum -2m^2x even the units do not agree (kg^2ms^-1) Tho attempt is good but if only someone could refine it to get a concise expression for t. Thanks

Just wondering if any one in left 19 hours can establish a mathematical relation in friction, mass ,force and time. (intertia is just a measure of mass)

Answer 1

F=MV

Answer 2

The affecting factors here are two : one, the frictional force and two, the center of gravity.

The process can be described as below:
The last coin experiences two forces when it is hit:
1) the downward force caused due to gravitational pull. It also has the center of gravity of the whole pile.
2) Due to the downward pull and contact, it experiences frictional force as well.

Now when u hit the last coin with enough force such that is removes the last coin overcoming the frictional force quickly enough that the center of gravity is tranfered onto the next coin without causing the applied force to affect the position of the rest of the coins, then the pile remains as it is.
If not, it disturbs the coins' position and instead of a single center of gravity each coin has a different center of gravity. Now, remember that an object can be only balanced at its center of gravity. Since there is no single place in this case, the pile falls.

Answer 3

Let's see if this works for you.

...a body at rest tends to stay at rest unless acted upon by an outside force.

If you hit the bottom coin fast enough it clears the stack before the force of gravity can overcome inertia and the whole stack falls straight down as a unit. Slowly remove the bottommost coin slowly and gravity acts upon the unsupported portion of the stack, tilting the stack, causing gravity to pull on the whole side of the stack and causing it to fall.

Answer 4

The answer to this actually has to do with friction. You must apply enough force on the penny to very quickly get it moving, which will break the hold of static friction between the bottom penny and the one above it. Once this happens, the force that you apply to the bottom penny that is transferred to the one above it is lessened, but there is still some. So, you have to apply enough force to very quickly pull out the penny, before the bottom penny moves out of alignment enough to cause a collapse. Also, you must move the penny fast enough so that by the time the side of the stack behind the moving penny has fallen to cause a collapse, the leading edge is moving as well to cancel the effect.

Answer 5

It gives the bottom coin the momentum to get "out from under the pile" in time for the coins to settle correctly in their alignment, in essence, you're only removing or touching the bottom coin.

Without this momentum, the bottom coin isn't given the speed to clear and knocks out of alignment the rest of the stack thus knocking most of them down due to inertia of the first coin striking and "not being out of alignment." The fact that they are not stacked has little to do with their falling, however, the fact they were "struck with force," does.

Answer 6

The force lateral applied on the bottom most coin has to be greater than the gravitational force of the other coins above so that before gravity starts to act the coin is moved out

Answer 7

Apart from points by Larry and Adam,
the speed with which we hit also gives it enough force (because of high acceleration) to over come friction between the coin and the above stack helping it get out from below, before other forces destabilize the stack

Answer 8

pile of (n+1) coins, each of mass m, hit with a coin of mass M and velocity X m/s

Friction between coin at the bottom and the one above it = u.n.m Newtons where u is the coefficient of friction and assume the stack is resting on frictionless surface.

impuse force min = F.t = u.n.m.t where t can be in milliseconds
chage in momentum before and after collissions = m.X - m(2x)

where x is the speed of both coins after collision in the same direction

u.n.m.t = m.X - m.(2x)

Answer 9

The normal reaction is altered marginally in its equillibrium condition if the pile is hit fast. On the other hand if it is hit slowly the applied force will disturb the equillibrium thereby causing the pile to fall in the direction opposite of the resultant force.

Answer 10

upper coins are in a state of inertia of rest.when you hit slowly to the lower coin,it compels the upper to move and hence they all fall.the same thing dosen't happen when you hit the lower coin faster.

Answer 11

And your coefficient of friction would be???

Nuclear bonding? Small matter attractors? Gravity? Corielos effect- yes I know it's spelled wrong- it makes my head spin...