How to make a boy fall from the merry-go-round?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\mu mg$.