# How is normal force in circular motion?

Let's say I have a vertical ring and a bead is sliding on it. I understand normal force is always acting radially and that in general the radial component of weight is not equal to the normal force. In fact the difference between these 2 forces is what causes circular motion. What I don't understand is what makes normal force is not equal to that component. I mean what is happening physically between the bead and the ring?

In general the normal force is pretty tricky. If I could give loose definition to the normal force it would be "the force needed to constrain motion along the surface the object is constrained to given the object's trajectory."$$^*$$

As you start digging into it, the normal force is pretty complicated. It is a force that depends on the position and velocity of our object as well as the other forces acting on the object. It isn't a force where we can just look at a "snap shot" of the object and know what the normal force is.

I mean what is happening physically between the bead and the ring?

The ring is exerting a force that keeps the bead constrained to it. This is because the bead cannot move through the ring.

More mathematically speaking, the radial net force component is given in general by $$F_{rad}=w_{rad}+N=m(\ddot r-r\dot\theta^2)$$

and since the weight will be given by $$w_{rad}=mg\cos\theta$$ ($$\theta=0$$ is at the bottom of the ring), and because on the ring $$\ddot r=0$$, we know that the normal force is what is "left over". i.e. $$N=-mg\cos\theta-mr\dot\theta^2$$

This is what $$N$$ needs to be too keep the bead on the ring (to make $$\ddot r=0$$)

In general things can be more complicated if $$\ddot r\neq0$$. For example, if you consider the classic banked curve problem but don't assume your object is moving around the same circle the entire time, you will find oscillatory solutions where $$r$$ is changing over the entire trajectory.

$$^*$$ Of course this seems like a circular definition, but you will find that this is the kind of thinking we get when you move to higher level mechanics like Lagrangian mechanics. We actually don't need to know what the normal force is to determine the trajectory of the object this way. You can find the trajectory using Lagrangian mechanics and then determine what the normal force needed to be in order to obtain that trajectory using Newtonian mechanics.

Any object for a circular motion, has a change in velocity,( atleast a change in direction!). This change or acceleration implies a force is acting on it. If you try to analyse the direction of change in velocity (in a small interval) using simple laws of vector addition you will find that the change takes place along radial direction, and it requires a radial force , towards centre. This is called centripetal force. In case of a bead, the normal contact force provides necessary centripetal force.
Consider bead sliding on horizontal ring first.At each instant, the tendency of bead is to move in the direction of its velocity (that is tangential to the ring). The bead tries to move out of the ring, so the inner surface of bead presses against the ring such that it tends to move outwards, but the contact generates a contact force inwards, which supplies the required centripetal force. In this case gravity doesn't contribute anything.
Now if we take ring to be vertical. Consider bead is slightly disturbed from rest from top of ring such that it starts sliding down due to gravity. Now the mg supplies the required centripetal force initially (which is small required as speed is small initially). So a normal reaction appears in radial direction away from centre such that: $$mg-N=mv^2/r$$ Now, as ring slides, it's speed increases. As it moves towards the vertical Portion of ring(after covering an acute angle with respect to initial vertical position), the centripetal force contributed by mg decreases ask now it's component $$mg cos\theta$$ instead provides the centripetal force. But now the required centripetal force increases , on the other hand. So the bead presses outwards and the normal reaction changes direction from radially inward to outward. Now $$mgcos\theta + N= mv^2/r$$ where \theta is angle covered by ring from upward vertical. It is general equation, but here v is also a function of theta(the decrease in potential energy causes increase in KE). At bottom you can see $$N-mg=mv^2/r$$ ($$\theta =\pi$$). The normal reaction is vertically upward. Again in its upward journey, the normal reaction changes direction from radially outward to inwards(that means normal reaction becomes zero for an instant). Hope my explanation helps you!