Understanding Centripetal Force: Free Body Diagram of a Roller Coaster Loop

[transmini]

So I understand that centripetal force is "center-seeking", however I'm curious as to how this looks on a free body diagram. I know there is no arrow on the FBD representing centripetal force. However, if we're looking at say a roller coaster at the top of a loop, gravity points downward, and normal force points downward. So centripetal force would end up being downward. How is it then they the coaster stays against the track? What force is opposing the force due to gravity and the normal force? Also in this particular scenario, what forces would be included in the centripetal force?

 

[Doc Al]

How is it then they the coaster stays against the track? What force is opposing the force due to gravity and the normal force?

The coaster is moving fast enough that it is pressed against the track. (That's why there's a normal force. If you go too slow, you fall off the track.) There is no force acting besides gravity and the normal force.

Also in this particular scenario, what forces would be included in the centripetal force?

The net force of gravity plus the normal force is the centripetal force.

 

[transmini]

But what would we call the force pushing against the car into the track? Or is just something kind of there? Also, on a previous problem I did with someone, for a jet circling in the air, the centripetal force was some number, and to find apparent weight, we added or subtracted the weight from gravity. Wouldn't that imply in these scenarios gravity isn't included in the centripetal force? Or are the scenarios different in this case?

 

[Doc Al]

But what would we call the force pushing against the car into the track?

The normal force. It is an interaction between the coaster and the track--they push on each other.

 

[Doc Al]

Also, on a previous problem I did with someone, for a jet circling in the air, the centripetal force was some number, and to find apparent weight, we added or subtracted the weight from gravity.

The apparent weight is just the normal force the passenger in the plane would feel pressing against him.

 

[IgorIGP]

I know there is no arrow on the FBD representing centripetal force.

Why is that? The centripetal force is the role played by one force or resultant force from the sum of several forces. The specifics of this role is that the presence of non-rectilinear movement always needs in such a force. In the different cases, for example rotational movement, this role can be played by various forces. But always it can be drawing as a vector (arrow on FBD). Must be said that the centripetal force is defined as a necessary and sufficient condition for curvilinear motion, and it is not limited itself by what forces it can be "made of".

 

[vela]

So I understand that centripetal force is "center-seeking", however I'm curious as to how this looks on a free body diagram. I know there is no arrow on the FBD representing centripetal force. However, if we're looking at say a roller coaster at the top of a loop, gravity points downward, and normal force points downward. So centripetal force would end up being downward. How is it then they the coaster stays against the track?

You're forgetting the law of inertia. If no forces were acting on the roller coaster, it would keep moving in a straight line. There's no force necessary for an object to continue its state of motion. The normal force and weight combine to alter the motion of the roller coaster from a straight line so that it follows the circular path. There's no force needed to hold the roller coaster up against the track.

If this still doesn't make sense to you, consider a projectile at the top of its parabolic trajectory. (You can argue the following about any point on the trajectory, but at the top is probably easiest to picture.) The only force acting on the projectile is gravity, which points straight down, yet the projectile doesn't fall straight down from the top of its trajectory. Why not?

 

[IgorIGP]

transmini, now I understand that you need to find out that:

How is it then they the coaster stays against the track?

When you are sitting on the chair, you press it by your weight and it press you by the reaction that is equal to your weight. So we have an equilibrium with two forces acting on you: the gravity and the chair reaction. That is a static equilibrium. The state of rest is the state when the velocity is equal zero. You are not moving because the forces acting on you compensate one each other. When you are jumping up the state at top point is a rest state, because it is a moment when the velocity is equal zero. This is because moving up you stop and then move down, ie the velocity changes its sign and then its value passes through zero. Note that the motion is due to the force is the force of gravity. Nevertheless, we have a stationary point in this case. Such a state is called as a dynamic equilibrium. vela shows a good exsample with the projectile. Jumping on a moving platform you can move as the projectile moves. The highest point of each jump is a very special point. This point differs from the others in that the force of gravity is perpendicular to the path and does not make work. Each such a point is the point of momentary rotation and gravity plays the role of centripetal force in that case. Imagine that any uniform rotation is a contiguous change of the simmilar equilibrium points.

 

[drvrm]

i wish to add that for the motion of a body on a curvrd path like roller coaster one needs a centrpetal force - as you are in a frame moving in a rotational manner about a centre-an axis passing through the centre; but suppose one may not provide it or sufficient force is not provided for -
in absence of such a force an equal but opposite force called centrifugal will act and that will press you on your seat the seat will give reaction and that reaction working inwards does provide the centripetal force - but its in a frame rotating about the axis-or call it accelerated frame.
to visualise it one has to be on the roller coaster.

 

[Doc Al]

i wish to add that for the motion of a body on a curvrd path like roller coaster one needs a centrpetal force - as you are in a frame moving in a rotational manner about a centre-an axis passing through the centre; but suppose one may not provide it or sufficient force is not provided for -
in absence of such a force an equal but opposite force called centrifugal will act and that will press you on your seat the seat will give reaction and that reaction working inwards does provide the centripetal force - but its in a frame rotating about the axis-or call it accelerated frame.
to visualise it one has to be on the roller coaster.

In order for something to undergo centripetal acceleration, centripetal forces are needed. These are "real" forces (with agents that exert the force) that exist in all frames.

When viewing things from a non-inertial frame, so called 'fictitious' forces are used so that Newton's laws can be applied. Centrifugal force is one such fictitious force. It only exists as a convenience when working in a rotating, non-inertial frame. Such "forces" do not exist when using the usual inertial frame of reference.

 

[IgorIGP]

Doc Al, I agree without talking.

. Such "forces" do not exist when using the usual inertial frame of reference.

It can be said even more strictly: "Such "forces" do not exist" and all.

 

[A.T.]

Doc Al, I agree without talking.

It can be said even more strictly: "Such "forces" do not exist" and all.

Forces in general are invented by humans for convienient calculations.

 

[IgorIGP]

Forces in general are invented by humans for convienient calculations.

Of course no. Force reflect the actions and they exist as such a reflection without our desires. A centrifugal forces do not exist due to lack of appropriate action. They really invented for ease of calculation. This is a convenient technical approach.

 

[drvrm]

i agree to the concept of 'fictitious' forces generated by non inertial frames.

 

[IgorIGP]

forces generated by non inertial frames

Inertial forces are never generated by any kind of frames. Inertial forces are generated by the human who needs to write the II law of Newton for non inertial frame. Such a record is physically incorrect, but it is mathematicaly correct due to that "fictitious" forces. Such an approach is justified by a sharp decline in the technical complexity of the mathematical description. But there is a deformation of physical picture take place.

 

[drvrm]

Forces generated in a non inertial frame means-those experienced by an observer in non inertial frame-apart from mathematical representation the non inertial description -say a man going up in a lift does experience those fictious ones- or a river flowing on Earth does bend towards the direction of coriolis force.

 

[IgorIGP]

a man going up in a lift does experience those fictious ones

No one can feel fictious force. Even in lift. That a man feels the force acting from lift to him and nothing mysterious or fictious. I feel I can not convince you, and I think the continuation is of pointless.