# A coin experiment

1. Mar 8, 2015

### Prannoy Mehta

A coin can stay upright on its rim for a some time when rolled with a some velocity, while it falls from its upright position at the slightest disturbance when stationary. I did not understand why it happened. I just dropped a few coins on the floor and noticed this thing which is very new to me.

2. Mar 8, 2015

### Stathis V

1. Angular Momentum
2. Mass
3. Center of Mass
4. Moment of Inertia

3. Mar 8, 2015

4. Mar 8, 2015

5. Mar 8, 2015

### Prannoy Mehta

I din't understand it completely. So the torque by the angular momentum cancels of the shift of centre of gravity. Causing no change. Where as when the body is not in motion, a slight disturbance of centre of mass coin causes it to fall.

6. Mar 8, 2015

### mr166

Sometimes a non-rigorous explanation is easier to understand. A rotating object resists twisting perpendicular to it's axis of rotation. Thus a rotating coin does not fall (twist) down. FYI that is also why a bicycle is easier to keep upright when it is moving. It is called the gyroscopic effect.

7. Mar 8, 2015

### A.T.

It doesn't "resists twisting perpendicular to it's axis of rotation". It changes the axis of rotation with an 90 offset to the applied torque. For the rolling coin this means that the tilting torque of gravity, actually steers the coin into a turn. The coin doesn't tilt further, because the centripetal contact force required for the turn also creates a torque which cancels the torque of gravity.

This is only relevant when you take the hands of the handles, so the bike has to stabilize itself. With hands on the handles you do the steering to avoid falling over.

8. Mar 8, 2015

### Prannoy Mehta

So, I think I have I understood it now. Thanks a lot :)

So here is what I have understood. There are three axis of rotation.
One, where the coin is taking a turn inside a circular path.
Second, where the coin rotates across its centre.
Third, the bottom of the coin where the axis is added when an external force is applied (Which can be assumed to act on the Centre of mass of the system)

Now, another question is, if it is balanced by the centrifugal torque right, because mv^2/r always acts outwards. (This is what I am learning, in my Physics classes, so please correct me if I am wrong) Now let the radius of the circular loop being carried out at one instant be at -x direction, so the centrifugal force will be acting in the x direction. Suppose I chose apply an external force(In the x direction, by blowing the coin slowly, it should be enough to turn the coin). The centre of gravity(COG/COM)will shift, causing the coin to again collapse by its self in the outward direction, as now two torques act outward. But some how they seem to be going inward as soon as the coin slows down. (And the radius decreases enormously. Why does that happen would be an added question)