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I'm whirling a ball tying it by a rope. Then at any instant the centrifugal force (outward force) equalizes the centripetal force (inward force) . Then both should cancel each other and there should be only tension of the rope to balance the weight of the ball. Then why do we count the centripetal force as resultant force. As an example in the mean position $T=mg + m\frac{v^2}{R}$ why do we also consider the second term?

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Suppose, you are analyzing the situation from the your frame of reference. Then, the net radial inward force on the ball is, $T - mg \sin \theta$, while the inward radial acceleration is $\frac{v^2}{r}$. Applying Newton's second law, $T - mg \sin \theta$ $=$$\frac{mv^2}{r}$.

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Now if you analyze the motion from the frame of the ball, there is no radial acceleration. So, the net radial force must be zero. The inward radial force is, again, $T - mg \sin \theta$, while there is the centrifugal force (pseudo force) along the outward radial direction having magnitude $\frac{mv^2}{r}$. So, we get, $\frac{mv^2}{r}-(T-mg \sin \theta) =0$. Thus, the equations are same in both of the frames.

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From the reference frame of you whirling the ball, there is no centrifugal force. The centrifugal force is a fictitious force, only showing up in the reference frame of the ball. This is because the ball's reference frame is not inertial (it is traveling in a circle, and therefore is constantly accelerating). Noninertial frames gives rise to fictitious forces like the centrifugal and Coriolis forces.

The tension is the centripetal force (it is what keeps the ball moving in a circle). So the only forces acting on the ball, from your reference frame, are the centripetal force (tension) and gravity (neglecting air resistance). There is no centrifugal force, in your frame, to cancel the centripetal. If there was a force cancelling the centripetal force, there wouldn't be circular motion.

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  • $\begingroup$ If that is true then we should write $T=mg$ always. $\endgroup$ – Nobody recognizeable Apr 28 '18 at 0:27
  • $\begingroup$ What is $T$? Tension? $\endgroup$ – roshoka Apr 28 '18 at 0:28
  • $\begingroup$ tension is a force which takes accelerates the ball towards the center. $\endgroup$ – Nobody recognizeable Apr 28 '18 at 0:32
  • $\begingroup$ $T$ does not always equal $mg$. Think about the direction of each force. Gravity is always downward, but tension is always pointed toward the center of the circle that the ball is moving in. So gravity and tension will not always be in the same direction. So you can't say $\vec{T}=m\vec{g}$ always. $\endgroup$ – roshoka Apr 28 '18 at 0:33
  • $\begingroup$ Are you spinning this ball horizontally or vertically? $\endgroup$ – roshoka Apr 28 '18 at 0:42
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The equation $$T=mg + {mv^2\over R}$$ holds only for vertical circular motion at the instant the ball passes through the lowest point of the circle. Gravity(Weight of the ball) is always pointed down and so will be diluted as the ball passes around the circle except for the bottom like I said earlier. The tension in the rope constantly pulls the ball in while it moves in a circle and that is the source of the centripetal force. That is to say that centripetal force should not be regarded as a different kind of force than the rope tension. It is this tension in the rope that we call the centripetal force. What we normally call centrifugal force does not apply to the ball but instead the rope itself. By Newton's third law, the ball exerts equal force(called centrifugal) on the rope that the rope pulls on it ( centripetal force). So the "centrifugal force" doesn't affect the ball but the rope. Hence, the term $mv^2/R$ is the force due to the acceleration that the ball experiences ($F=ma$ remember). But it is the tension $T$ (centripetal not centrifugal) that is causing this while carrying the weight ($mg$) too. Effectively, at the very bottom of the vertical circle, $T=mg +mv^2/R$. It is good to note that the term $mv^2/R$ is neither the centripetal nor centrifugal force. It is the force due to the acceleration that the ball experiences. The centripetal force is the sum of this term and the weight. This centripetal force is same value as centrifugal force(by Newton's third law).But I repeat, the centrifugal force is on the rope not the ball so they can't cancel each other like people assume. Hope that one helps

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  • $\begingroup$ Kk tank u sir I have went through it $\endgroup$ – EMMANUEL CHIDERA May 1 '20 at 22:39

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