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I want to make this boy fall in the game and not able to figure out a way or formula which is responsible to make this boy fall from merry go round

I want to make this boy fall in the game and not able to figure out a way or formula which is responsible to make this boy fall from merry go round

We know centrifugal force and centrifugal force theory. When a boy standing on merry go round he will experience a centripetal force along with centrifugal force. My question is when & why boy will fall from the merry go round? How to calculate centrifugal force because of this force boy will fall? What is the condition for calculating force acting on a boy when he falls from merry go round? and if there is no centrifugal force then on what basis i should make the boy fall from the marry go round?

Community
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Kartik Shah
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    Related: https://physics.stackexchange.com/q/109500/2451 and links therein. – Qmechanic Dec 17 '18 at 09:35
  • You should add a little more information to your question to make it more clear. If I'm not wring, your guy is moving towards the outside of the merry go round (as shown by green velocity arrow) while the its rotating? Also, the merry go round's rotation is accelerating (the rotation is speeding up as given by black acceleration arrow)? How is he holding on to the merry go round? is he using his hands and feet? Are you asking how fast the merry go round has to go before he can't hold onto it anymore and falls off? – Skawang Dec 18 '18 at 13:18
  • @Skawang Ignore the arrow position and my question is how the boy will fall from the merry go round? or what force is behind falling of that. But solved it using friction. (μmg) – Kartik Shah Dec 19 '18 at 10:47

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First, be aware that there is no such thing as a centrifugal force. The "centrifugal force" is not a force! It is just a feeling. A feeling that is a result of the body's tendency to continue moving due to inertia. There is no "formula" for a "centrifugal force" as such, because it doesn't exist. It just feels like it.

What does exist is this tendency to continue moving straight when in motion. And "moving straight" on a merry-go-around or any other circular motion (a turning car for instance) means moving away from the centre. In order to not move away from the centre and instead follow the merry-go-around around the circle, a centripetal force must be exerted on you inwards. You mentioned that in the question. But there is no "centrifugal force" outwards at he same time. It is just the body resisting to move along with the centripetal force. I tend to call it a centrifugal effect, to have a better word for it than "centrifugal force", since it isn't a force.

If nothing stops you (if you don't hold the rail of the merry-go-around), then you will continue moving straight away from the centre and there is no force on you that does this (there will only be a friction force from the ground when you eventually fall).

But if you hold the rail or are being blocked by the rail or similar, then the rail pulls you along with it. Pulling you along is done by a force that depends on the situation:

  • If you lean against the rail, then the rail exerts a normal force on you which gives you the necessary radial acceleration that causes you to follow along in the circular motion.
  • If you hold the rail then the force might come from your arms.
  • In a turning car, you feel squeezed into the door which holds back with a normal force or pulled by the seat belt.
  • Etc.

This force is the centripetal force as you mention. And this is all there is. A centripetal force and then a radial acceleration. The centripetal force has no specific formula, because it depends on the situation as described. The radial acceleration does have a formula, as you may already know:

$$a_r=\frac{v^2}r$$

But we cannot get any more specific. There is no such thing as a "centrifugal force"; it is not a force which we can set up a formula for.

Steeven
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  • I appreciate your reply, but my question is not resolved yet. I want to make boy fell from the merry go round, which formula lie behind that effect? the formula which is pulling a boy inside a circle is a centripetal force (as per my knowledge) but I am not able to figure out when the boy will fall from the merry go round? and what formula or physic rule act behind falling of the boy? – Kartik Shah Dec 17 '18 at 11:32
  • @KartikShah He will not fall from the merry-go-round unless the centripetal force stops working. So it all depends on that force. In order for him to swing around, he must have a certain inwards radial acceleration (see the formula in the answer). Such acceleration requires a certain force, the centripetal force. If it is not possible to exert a large enough centripetal force to achieve the radial acceleration, then he will fall out. If he holds on to the rail and his arms cause the centripetal force, then maybe at some point he can't hold on any longer because the needed force is too large. – Steeven Dec 17 '18 at 11:37
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    @Steeven I don't agree with your strict statement there is no such thing as a centrifugal force. Apparently you only accept to do physics in an inertial frame of reference. But there are many cases when a non-inertial frame is useful or more natural, and it's there (you surely know it very well) that a centrifugal force appears. As an instance, we live on earth which is rotating. Because (partly) of this, gravity acceleration is less at equator than at poles. Would you refuse to reason in Earth's frame? And in your line of thought, what about Coriolis force and meteorological phenomena? – Elio Fabri Dec 17 '18 at 16:02
  • @Steeven I would like to add something to your answer. Merry go round is already revolving. My question is not to revolve merry go round. My point is if merry go round is revolving then at what cases the boy will fall.

    If the centripetal force stops working then merry go round will stop. So why the boy will fall when there is no revolution.

     – Kartik Shah Dec 18 '18 at 05:36
  • @ElioFabri You are welcome to do physics in non-inertial frames. To do that you'll have to "invent" the centrifugal force in order to keep Newton's laws working. I am not saying that the use of such fictitious forces and the use of non-inertial frames shouldn't be done - just that we should be aware that it indeed is a fictitious force. The only reason that I object to the use of the term "centrifugal force" in general is that it causes so much confusion – Steeven Dec 18 '18 at 07:06
  • @KartikShah I am sorry, I don't fully understand what you mean. Would the boy fall even without any rotation? I think I need to see a sketch to fully understand what you mean. – Steeven Dec 18 '18 at 07:08
  • @Steeven Could you please explain the condition to study the cases of boy falling form the merry go round. I would appreciate if you could help me with the calculation part because I have studied the theory but I am unable to get the exact mathematical model to study the cases of a body falling from merry go round. – Kartik Shah Dec 18 '18 at 10:40
  • @Steeven I have added a graphical representation for your understanding. I hope this helps. – Kartik Shah Dec 18 '18 at 12:48
  • @KartikShah "explain the condition to study the cases of boy falling" If the merry-go-round is spinning, then it has a certain angular acceleration $\alpha$. Depending in where the boy is located, he needs a certain (radial) acceleration $$a_r=r\alpha$$ A force must cause this acceleration. Maybe a normal force if he is leaning against a rail, maybe friction on the floor, maybe a force from his arm if he is holding on. Set up Newton's 2nd law and include this force $f$ $$\sum F=ma\quad f=ma_r$$ The force must be strong enough to fulfil this equation. Otherwise he will be swung away. – Steeven Dec 18 '18 at 16:54
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The first thing I want to tell you is that centripetal or centrifugal or not different kind of forces they are just resultant forces so don't treat them like different kind of forces, they are just resultant forces acting towards the center in case of circular motion.

Aziz Lokhandwala
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    Nomenclature-wise "centrifugal force" is not the same as centripetal force. It has it's origin in the centripetal force but they are different critters. Centripetal force is whatever combination of actual forces are creating a curved trajectory as observed in an inertial frame. The centrifugal pseudo-force is one of several mathematical corrections added to allow you to treat a non-inertial frame using the usual tools and techniques of inertial mechanics. – dmckee --- ex-moderator kitten Dec 18 '18 at 02:18
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As pointed out, centrifugal force is imaginary force. It is just a reaction force to centripetal force, which acts towards the centre. The boy only falls off if the reaction force to centripetal force, $m\frac{v^2}{r}$ is greater than what is holding the boy back, which is friction, defined by $\mu mg$.

QuIcKmAtHs
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