# Confused in diffferent forces involved in circular motion

1. Nov 20, 2014

### shaks

Hi,

I am trying to solve one problem of circular motion and I searched on Internet and found many things and I am confused about different forces or I think same force is shown as different symbol. What I learned is as follows:

Force = F = ma (horizontal movement - where height is not involved)
Force = F = mg (vertical movement- where height is involved -circular/rotational motion)

Centripetal Force = Fc = m*ac (circular or rotational motion)
Normal Force = Fn or T or N = Fc + mg (circular or rotational motion)
Normal Force = Fn or T or N = Fc - mg (circular or rotational motion)

Sometimes there is also Fnet, Fg used. Or might be different site or man explaining uses different symbol for same force. Can someone please explain. Fg is I think gravity force and that's already calculated somewhere like mg or m*ac. So other than standard gravity force and frictional force I am confused.

Mine problem.
I want to calculate which forces are working on roller coaster at top and bottom in 100% unified circular motion. I am not student but my project is exactly like roller coaster i.e. object/mass moving in vertical circle. I want to calculate how much net force is acting on top and how much on bottom. So can find power according to force value.

Shaks

2. Nov 20, 2014

### Simon Bridge

You have seen the force calculated for different circumstances. The different symbols are just notation.
The definition is from Newton's second law: $\vec F = m\vec a$
... notice that both force and acceleration are vectors? This is key.

So gravity acts straight down in proportion to the mass, so $\vec F = mg \text{<down>}$ (which is also called "weight") ... notice that force has a magnitude and a direction. This may be called $F_g$ or Bob or Sue or anything that takes the author's fancy. The label is not as important as the concept.

If you want to lift something against gravity, at a constant speed, then you must apply an equal and opposite force during the motion... the $\vec F= -\vec F_g = mg\text{<upwards>}$ Notice that I had to use $F_g$ for gravity here, so that I can distinguish the force of gravity from the applied force of the human doing the work.

The total force on the object is $\vec F + \vec F_g = (mg-mg)\text{<upwards>} = 0$
... for a constant speed, the total force must be zero, non-zero forces make acceleration.

... for something this simple, the direction is often left to the context. i.e. if someone says the weight is 100N, you don't need to be told that the direction is downwards.

Velocity is a vector too. You can change velocity by changing speed (the magnitude) or by changing direction.
If a mass m turns a corner but maintains the same speed, then it's velocity changes ...
A changing velocity is an acceleration so there must be an unbalanced force.

The condition for an object to travel on a circular path is that the force has to point to the center of the circle, and it has to have a magnitude related to the speed that it is travelling and the radius of the circle. But it is still F=ma. It is always F=ma. If the force is different from that condition, thenthe motion just won't be in a circle. It'll be a spiral or something.

3. Nov 20, 2014

### shaks

Thanks for replying and trying to explain these basics to me. Can you please answer these questions.

1. When we are calculating something on horizontal movement then we say w=mg (Weight = mass*gravity) and when we are calculating something on vertical movement then we say F=mg (Force = mass * gravity). It means both are same we say it "w" or "F" or "Force" or "Fg" its same, right? Just due to type of motion we use different notations otherwise its same thing that's force. Weight is also a force that in physics terms.

2. In circular motion "net force" = Fg (mass*gravity) + Fc (centripetal force), right?

Shaks

4. Nov 20, 2014

### Simon Bridge

None of the above are strictly correct because force is a vector - so the direction must be included somehow.
When those equations get used, the direction is usually given in the context - which you have excluded above, resulting in confusion.

When we talk about moving vertically, we are usually referring to an applied force - i.e. not from gravity. If the motion is vertical at a constant speed, then the applied force has to be equal and opposite to the weight, so we'd say that F = -w = -mg<down> = mg<up> (since <up> = -<down>).

This is why different labels may be applied in different circumstances: to distinguish them. It is important to distinguish between the force of gravity and the force of some other object that just happens to have the same magnitude.

Further, vertical motion need not be at a constant speed, in which case F > mg<up> would mean acceleration upwards and F < mg<up> would mean acceleration downwards.

No. The net force is the sum of all the forces on an object. The component of the net force which points towards the center is called the centripetal force, the rest is called the tangential force.
So $\vec F_{net} = \vec F_c + \vec F_t$
In circular motion, the net force must point to the center of the circle ... so, $F_t = 0$.
In uniform circular motion, $F_c=mv^2/r$

(Note: last two equations are magnitudes, the directions are implied by the context.)

Last edited: Nov 20, 2014
5. Nov 20, 2014

### shaks

Thank you man, I have learn something by above post.

One thing that is "tension", tension is only involved when object is in unified circular motion but tied with cable, rope, rod, etc. In roller coaster example, tension is not involved? In simple words that leads to my project I want to calculate how much force is working (i.e. required) at bottom to move object in unified circular motion. Fc = mv2/r I already know I want to calculate net force required?

Shaks

6. Nov 20, 2014

### Simon Bridge

Tension is not restricted to cables, but anything which is pulled on at each end.
In real engineering you also have to deal with stress and strain and shear etc.

For a coaster going around a loop - the force needed to make the motion uniform will vary with the position on the loop. At the bottom of the loop, you need only what little force is needed to exactly match friction - in idealized frictionless model that would be zero. The normal reaction force from the tracks already points to the center.

7. Nov 20, 2014

### shaks

Clear

Can you tell me how to do this? My aim is to calculate power required to move object?

8. Nov 21, 2014

### Simon Bridge

Decide on the speed around the track - use this to work out the vertical component of velocity as a function of time.
The power needed at time t is then the instantaneous rate that gravitational potential energy is changing ... (+ friction).