Coin Experiment: Upright on Rim, Why?

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A coin can maintain an upright position on its rim while rolling due to the effects of angular momentum and gyroscopic motion, which help it resist toppling. When stationary, even a slight disturbance causes it to fall because the center of mass shifts without the stabilizing forces present during motion. The discussion highlights the role of torque and the interaction of forces acting on the coin, including centrifugal torque and gravitational torque, which balance each other during rolling. The gyroscopic effect is also noted as a reason why bicycles are easier to balance when in motion. Understanding these principles can clarify why a rolling coin behaves differently than a stationary one.
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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.
 
Physics news on Phys.org
1. Angular Momentum
2. Mass
3. Center of Mass
4. Moment of Inertia
 
http://megaanswers.com/unlike-a-stationary-one-why-does-a-rolling-coin-not-fall-down/

We consider the general case of a coin rolling on the floor with an inclination. The trajectory, in this case, is a curved path. This motion is a combination of three rotational motions each associated with a torque, one about its axis and the other about the line joining the centre of the coin and the centre of the curved path on the floor.
The combined effect of these torques counter-balances the centrifugal torque and the coin is automatically steered. Any additional toppling displacement is known as gyroscopic precession.

During the rotational motion of a circular disc about its axis, if a transverse torque is applied on it, the disc would simultaneously experience a turning torque about its third axis. Such a motion is known as gyroscopic motion that steers a rolling disc on to a balancing path. This behaviour ensures that the angular momentum of the rolling disc is conserved, a requirement of mechanics. Thus the coin does not fall.
 
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.
 
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.
 
mr166 said:
Sometimes a non-rigorous explanation is easier to understand.
But often wrong or misleading.

mr166 said:
A rotating object resists twisting perpendicular to it's axis of rotation.
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.

mr166 said:
FYI that is also why a bicycle is easier to keep upright when it is moving. It is called the gyroscopic effect.
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.
 
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 itself 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)
 
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