It is noteworthy to quote a sentence from my book,

It is a misconception among the beginners that centrifugal force acts on a particle in order to make the particle go on a circle. Centrifugal force acts only because we describe the particle from a rotating frame which is non-inertial.

Yes, the statement is undoubtedly right. But one thing that is annoying me is that If the particle moves on a circle due to centrifugal force from a rotating frame, what is the cause for motion of the particle on a circle from an inertial frame ? If there were only centripetal force , the particle would go towards the centre and never it would move in a circle. So, in order to move the particle on a circle , who will play the role of centrifugal force when viewed from inertial frame?? Please help.

  • $\begingroup$ That is right, centripetal force acts to force the circular motion. centrifugal force is just the reaction force to this. $\endgroup$ – ja72 Nov 15 '14 at 6:39

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 velocity is $a = v^2/r$. Remember that acceleration is just the rate of change of velocity, and it is velocity that dictates the direction the object is travelling. The centripetal acceleration is really only changing the direction of the velocity vector (not the magnitude) such that the velocity is always in a direction tangential to the circular motion.

Satellites for instance, are in constant free fall (at centripetal acceleration g), but it is moving fast enough tangentially that it misses the ground as it is falling towards the earth, the reason that its orbit is stable.

There is a "reactive" centrifugal force in the inertial frame, but it is a reactive force (per Newton's Third Law). It is applied by the particle to the object applying the centripetal force and is equal in magnitude and opposite in direction to the centripetal force. An example would be if one person was spinning another person and the two are held together by a rope, the tension in the rope is equal to the reactive centrifugal/centripetal force.

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    $\begingroup$ "moving fast enough that it misses the ground" - worthy of Douglas Adams (of Hitchhiker's Guide fame), and worthy of an upvote. $\endgroup$ – Floris Nov 14 '14 at 20:15
  • $\begingroup$ @Yandle: Oh! Sir, then if suddenly the centripetal force becomes zero, observer of inertial frame will see the particle to eject tangentially but the observer on the particle will see it ejecting radially?? So, how can it be ?? Which one is true??? $\endgroup$ – user36790 Nov 15 '14 at 6:32
  • $\begingroup$ @Yandle: I know what I see from inertial frame is true but I can't visualize how one from the particle can see the tangential ejection to be occuring radially! $\endgroup$ – user36790 Nov 15 '14 at 6:40
  • $\begingroup$ @user36790 I'm not sure what you mean when you say that the observer on the particle sees the particle eject radially. At the point where $a_c = 0$ (say the string snaps), the particle's frame is no longer non-inertial (as there's no net force on the object), if the particle leaves at v tangentially (relative to the center of rotation), the center of rotation should move at -v from the particle's perspective. Just to add that both the observations made in an inertial frame and non-inertial frame are "true", the observations are different, but equally valid. $\endgroup$ – Yandle Nov 15 '14 at 21:38
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    $\begingroup$ @user36790 The reason we have a "parabolic" path in projectile motion is because the horizontal velocity is not high enough to maintain a circular orbit. If you were to throw a rock (say on an airplane) at a v such that v^2/r = g, the rock would form a stable orbit around Earth (neglecting air resistance etc.) much like a satellite at r from the center of the Earth. If the rock loses velocity then r will decrease until it hits the ground, which would be the "parabolic" path you see in projectile motion. $\endgroup$ – Yandle Nov 18 '14 at 1:19

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 string and writes Newton's second law as $T = mv^2/r$, $T$ being the tension in the string, directed toward the center.

Now consider an observer in a frame of reference in which the stone is at rest. (Imagine him riding the stone.) He too observes the taut string. But he is at rest. To save Newton's second law, he imagines a force $F$ that exactly balance $T$ so that $F + T = 0$. This fictitious force $F$, is the centrifugal force.

As you see the observer in the inertial frame does not need $F$. It is contrived by the observer in the non-inertial frame so that he can still use Newton's second law.

Although Newton's second law is saved, Newton's third law is not. For in the non-inertial frame, the string pulls the stone with a force $T$ and as a reaction the stone also pulls the string by an equal and opposite force. While the stone experiences a centrifugal force $F$, no one experiences $-F$.


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