# Tag Info

0

The "rotating particle" picture does not work for $L=0$, unfortunately. Both the electron and the nucleus are waves who create a standing spherically symmetric wave of their relative motion.

1

It would be called centrifugal force. By definition you call it centripetal when it it is towards the center, and centrifugal if it acts opposite to it. If you have a centripetal force with an object in orbit, and suddenly you make it centrifugal (repulsive), the object will fly off, but I cannot tell if it will be an hyperbolic path without making the ...

1

Centripetal forces are actually just a consequence of the actual forces which act on a body, for instance the centripetal force which a body experiences when it is in orbit arises due to the fact that gravitation is an attractive force! So possibly for the centripetal force to operate in a opposite direction gravitation should be a repulsive force! (For ...

1

Let's consider a simple experiment in which a stone tied to a string is moving in a uniform circular motion in a horizontal plane. We can analyze this experiment from inertial and non-inertial frames. An observer in an inertial frame sees the stone having a radial acceleration and concludes that there must be a radial force causing it. He observes the taut ...

4

In an inertial frame the only force that causes a particle to move in a circular motion is the centripetal force. The reason that a particle does not "fall" into the center is because it has some tangential velocity, so it moves away from the center tangentially as it is falling towards it. The relationship between the centripetal acceleration and tangential ...

0

For an intuitive description, let's go to a semi-classical picture of the electron movement in the atom. I order to have an angular momentum, the electron should rotate around an axis of symmetry. Then, if in the plane perpendicular to the axis of symmetry the electron rotates, say, clockwise, the angular momentum will point along that symmetry axis. But, ...

-2

g-force is apparent weight/true weight therefore g-force is ma+mg/mg .

1

First rewrite the equation as $R_2 - mg\cos \theta = \frac{mu^2}{a}$. The reason the signs are different is that here $R_2$ should point towards the center, unlike in your drawing. Gravity points outwards and so it appears with a minus sign.

-2

The answer is obvious by inspection. Unfortunately, researchers don't bother examining the object of study but create math models with assumptions that drive the result. Centrifugal force driven by the bicycle steering into alternating arcs is the most common assumption. Then there's the utter arrogance of physicists who won't study tire forces. It is an ...

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