Understanding centripetal force and free body diagram

[jsmith613]

I don't like the way you used speed [I underlined it]!
They do accelerate at the same rate [speed is not a term that relates well to acceleration]. The fact is that acceleration is greater than g, so the only way the water can accelerate at a rate greater than g, is if something applies an extra force to it [gravity is not enough] The thing that applies the extra force is the bucket - but that can only happen if the water remains in contact with the bucket - it doesn't come out.

What I still don't get is that the water is accelerating downwards at the top of the flight, no?

 

[PeterO]

What I still don't get is that the water is accelerating downwards at the top of the flight, no?

Anything traveling in circular motion is accelerating towards the centre of the circle. If the circle is a vertical circle - as in this case - the centre is down, when the object is at the top of the circle.
Just because an object accelerates in a particular direction doesn't mean it is going to travel in that direction.

If you are cruising in a car, then apply the brakes - you will be accelerating towards the direction you have come from, but merely reduce the rate at which you continue the way you were going. [If you take on that [negative?] acceleration for long enough you will stop and start going back where you came from.

A pendulum does that - in fact it is the strangest thing for when it is traveling at its maximum speed it has minimum acceleration, and when it is traveling at minimum speed [at the extremes], it has maximum acceleration.

 

[jsmith613]

Anything traveling in circular motion is accelerating towards the centre of the circle. If the circle is a vertical circle - as in this case - the centre is down, when the object is at the top of the circle.
Just because an object accelerates in a particular direction doesn't mean it is going to travel in that direction.

If you are cruising in a car, then apply the brakes - you will be accelerating towards the direction you have come from, but merely reduce the rate at which you continue the way you were going. [If you take on that [negative?] acceleration for long enough you will stop and start going back where you came from.

A pendulum does that - in fact it is the strangest thing for when it is traveling at its maximum speed it has minimum acceleration, and when it is traveling at minimum speed [at the extremes, it has maximum acceleration.

thanks for this :)

 

[jsmith613]

ok so based on this:

If I am on a roller coaster and I react a point such that my vertical reaction force becomes zero and I feel weighless, what expression can I write for mg in terms of mv^2/r and why?

http://www.edexcel.com/migrationdoc...E January 2010 - QP/6PH04_01_que_20100128.pdf

Question 12e)

So my free body diagram would just have mg down.
BUT I am also accelerating downwards with mv^2/r
so mg = mv^2/r

is this correct?

JUST TO SAY: the mv^2/r force will actually be at a slight angle to mg from the diagram, no?

 

[PeterO]

ok so based on this:

If I am on a roller coaster and I react a point such that my vertical reaction force becomes zero and I feel weighless, what expression can I write for mg in terms of mv^2/r and why?

http://www.edexcel.com/migrationdoc...E January 2010 - QP/6PH04_01_que_20100128.pdf

Question 12e)

So my free body diagram would just have mg down.
BUT I am also accelerating downwards with mv^2/r
so mg = mv^2/r

is this correct?

JUST TO SAY: the mv^2/r force will actually be at a slight angle to mg from the diagram, no?

Yes you are correct, the only force acting is mg, and the only force needed is mv²/R They will be equal in both magnitude AND direction.

The forces acting on you on a roller coaster are your weight [mg] and the Reaction force from the seat.
If your motion is such that the reaction force has magnitude zero, then the only force acting on you is mg.

As you go over the top of a hill on a roller coaster, you will be traveling in part of a circle - even if the hump is actually parabolic, the actual top of the hill will follow the path of a some circle.
While traveling in that circle, you have need of a centripetal Force of size mv²/R

Lets say F_c is 300 N [down], and your weight is 650 N. The reaction force from the seat must be 350 N [UP] to achieve that. You would feel light since before the ride started the reaction force from the seat was 650 N.
If you were traveling faster, F_c might be 500 N, so the reaction force from the seat would be only 150 N. You would feel even lighter.
If you were traveling faster, F_c might be 650 N. There is now no need for a reaction force at all. As a result you would feel weightless.
If you were traveling faster, F_c might need to be 700 N. Now you have a problem!
If this is a modern Roller coaster, which has 2 sets of wheels - one above and one below the "rail" - it is capable of generating a downward force. If the "safety Bar" has been correctly fitted - it will transfer sufficient downward force [50 N] to supplement gravity so you stay on the track.
If this was one of the old-fashioned Roller Coaster - you are in big trouble as you and the cart have just become a projectile.

 

[jsmith613]

Lets take a look at the 700N example.

I have 650N down (weight) and a reaction force of -50N
How can my reaction force be in the same direction as my weight?
this is odd :S

(forces have always baffled me, quite literally)

 

[PeterO]

Lets take a look at the 700N example.

I have 650N down (weight) and a reaction force of -50N
How can my reaction force be in the same direction as my weight?
this is odd :S

(forces have always baffled me, quite literally)

As I said - it has to be a new design Roller Coaster

http://www.madehow.com/Volume-6/Roller-Coaster.html

have a look at part 6 and you will see that modern roller-coasters have double wheel set-ups [above and below a tubular rail] so that as well as the standard upward force, they can supply a downward force if necessary. That is why modern roller coasters have a safety bar - to enable a downward force to be transferred to the occupants if necessary. It is also why they don't allow "little" people to ride, since the "safety bar" leaves room for small people to slip out from under.

 

[jsmith613]

no, what I am asking is how can the reaction force be -50 (i.e: in the same direction as weight).
As you said the Fc = 700N and mg = 650N so reaction force must be -50N
how does this make sense?

 

[jsmith613]

do you see my problem?

 

[PeterO]

no, what I am asking is how can the reaction force be -50 (i.e: in the same direction as weight).
As you said the Fc = 700N and mg = 650N so reaction force must be -50N
how does this make sense?

A reaction force is exactly that: a reaction to the situation.

When the cart goes slowly, or is stationary, the weight of the cart tries to make it sink through the structure, towards the Earth [down]. The wheels of the cart contact the rails, and the rails react by supplying sufficient upward force to prevent that happening - works fine if the rails/wheels are strong enough.
In the scenario where Centripetal Force needs to be 700 N, the cart will be trying to fly off the track. In a modern roller coaster, this will cause the lower set of wheels to push up on bottom of the track - with the reaction that the track pushes DOWN on the cart.

Did you look at the picture of a modern roller coaster - where there is a set of wheels above the rail as well as a set of wheels below the rail?
That is also why modern roller coasters can go upside down!
If the cart is moving as expected, the cars will remain in contact with the rails anyway. However, if someone's jacket fell out and became entangled with the wheels the cart may slow down too much and the cart would fall off the track [down] from the inside of an inverted loop. That possibility MUST be catered for, so the modern roller coaster has a complex wheel arrangement to prevent that. This arrangement also enables the coaster to pass over a hill at a speed which was previously too fast to remain on the track.

 

[PeterO]

do you see my problem?

I think so. The reaction force is not a reaction to mg, it is a reaction to what ever is trying to happen.
When you stand on the floor, the reaction force from the floor is equal in magnitude and opposite in direction to your weight, but does not constitute an action-reaction pair in the Newton's 3rd law sense.

The 3rd law pairs, or couples as they are often called, are.

The Earth pulls you down [with a force of mg down] - you pull the Earth up [with a force of mg up]

Because the Earth is pulling on you we also get;

You push down on the floor [with a force of size mg] - the floor pushes up on you [with a force of size mg]

When we do a free body diagram, we look at those 4 forces, and note that two of them are acting on YOU [the other two act on the Earth and the floor] and do a little picture representing you and those two forces.

We call one of those forces weight - and often give the tag mg to it.
We call the other force the Reaction Force - and often give it the tag F_R.
We sometimes use the tag F_N to emphasise that the force is perpendicular to the surface and is called the Normal Reaction Force.

**** that term Normal refers to "perpendicular", it is not some reference to the situation being common or the usual situation.

 

[jsmith613]

this partly cleared upon the netwon pair

What is confusing me is that I know Force = mv^2/r BUT this force is produced by other forces.
In our example (exam paper) the roller coaster is above the track so if we assume this is an old type (just for the sake of the question) what is providing the extra 50N to lead to a 700N force given the weight is constant?

is that clearer??

(by the way thanks a lot for all the help you've been giving me! it means a lot :)

 

[Doc Al]

What is confusing me is that I know Force = mv^2/r BUT this force is produced by other forces.

Best to think of mv^2/r as being the required net force to produce the given circular motion.

In our example (exam paper) the roller coaster is above the track so if we assume this is an old type (just for the sake of the question) what is providing the extra 50N to lead to a 700N force given the weight is constant?

Nothing! Meaning that if you go fast enough that you require a net force of 700 N to maintain contact with the track and you only weight 650 N then you cannot maintain contact with the track. You're going too fast!

 

[jsmith613]

oh so I require a force of 700N to keep me in a circle IF i am moving at v^2 = (700*r)/m
if there is not enough force to provide this I will not stay in a circle and will follow a tagental line of motion

thanks!

 

[Doc Al]

Right!

 

[jsmith613]

oh my gosh, I get it!
thanks so much everyone for helping me. I am sooo grateful for everything :)

 

[jsmith613]

ok, so one last summary check:

- if the force required to keep the object moving in a circle is not met by the forces acting on it, the object will not move in a circle (it will move in a parabolic path)
- if the force required to keep an object moving in a circle is met or exceeded the object WILL move in a circleso at the top of a loop-the-loop the only force acting are its weight (assuming it is a vertical circle and you are on the inside - and it is an old fashioned roller coaster).
Therefore, for such a roller coaster, it MUST be traveling at a speed that is equal (or less) to sqrt(gr) as otherwise the centripetal force required will be too large and not copmensated for by any other force?)

here's my reasoning:
if I go faster then the force required to keep me in a circle is NOT met by the forces acting on me
if I go slower then the forces acting on me are big enough to keep me in a circle
therefore, to move in a circle
v <= sqrt(gr)

(this contradicts the formula I found which is wierd)

 

[Doc Al]

ok, so one last summary check:

- if the force required to keep the object moving in a circle is not met by the forces acting on it, the object will not move in a circle (it will move in a parabolic path)
- if the force required to keep an object moving in a circle is met or exceeded the object WILL move in a circle

Careful with that second statement. If something is moving in a circle (at constant speed and a given radius) then the net force on it is exactly given by mv^2/r. (In the example of the roller coaster the 'reaction' force of the track on the coaster will adjust itself to be whatever needed to make that work out, as long as you don't go too fast.)

so at the top of a loop-the-loop the only force acting are its weight (assuming it is a vertical circle and you are on the inside - and it is an old fashioned roller coaster).

I think you mean on the outside. (On the inside of the loop is different, since the 'reaction' force points in a different direction.)

Therefore, for such a roller coaster, it MUST be traveling at a speed that is equal (or less) to sqrt(gr) as otherwise the centripetal force required will be too large and not copmensated for by any other force?)

Good!

here's my reasoning:
if I go faster then the force required to keep me in a circle is NOT met by the forces acting on me
if I go slower then the forces acting on me are big enough to keep me in a circle
therefore, to move in a circle
v <= sqrt(gr)

Good.

(this contradicts the formula I found which is wierd)

Perhaps you're mixing it up with being on the inside of the loop at the top. In that case, there's a minimum speed to maintain contact with the track--anything less and the coaster loses contact.

 

[PeterO]

this partly cleared upon the netwon pair

What is confusing me is that I know Force = mv^2/r BUT this force is produced by other forces.
In our example (exam paper) the roller coaster is above the track so if we assume this is an old type (just for the sake of the question) what is providing the extra 50N to lead to a 700N force given the weight is constant?

is that clearer??

(by the way thanks a lot for all the help you've been giving me! it means a lot :)

As Doc Al says, the old type Roller Coaster will not remain in contact with the track - only the new style with the compound wheel arrangement could manage that.

 

[jsmith613]

first are all your responses based on the roller coaster being on the inside or outside? (i.e: "Good")

I think you mean on the outside. (On the inside of the loop is different, since the 'reaction' force points in a different direction.)

On the inside of the loop at the top THERE IS NO reaction force as the cart is just about not touching the track, then it moves and makes contact again (this again assumes we are using an old roller coaster). Surely my statements work for this to be true

Perhaps you're mixing it up with being on the inside of the loop at the top. In that case, there's a minimum speed to maintain contact with the track--anything less and the coaster loses contact.

if you go at a speed greater than sqrt(gr) then the centripetal force required is not provided by anything other than gravity. How is this possible?

 

[Doc Al]

first are all your responses based on the roller coaster being on the inside or outside? (i.e: "Good")

Your comments made sense for the roller coaster being on the outside of the loop. (The coaster is right side up at the top of the loop.)

On the inside of the loop at the top THERE IS NO reaction force as the cart is just about not touching the track, then it moves and makes contact again (this again assumes we are using an old roller coaster). Surely my statements work for this to be true

On the inside of the loop, if you're going at exactly the right speed--the minimum speed--at the top, then there would barely be any contact force between coaster and track. You can of course go faster to maintain good contact with the track. Go slower and you fall. (The coaster is upside down at the top.)

if you go at a speed greater than sqrt(gr) then the centripetal force required is not provided by anything other than gravity. How is this possible?

Again, you seem to be mixed up between outside and inside the loop. Here's an illustration of the coaster being inside the loop:
[URL]http://www.physicsclassroom.com/mmedia/circmot/rcd.gif[/URL]
See the discussion here: http://www.physicsclassroom.com/mmedia/circmot/rcd.cfm"

 

[jsmith613]

doesn't going faster mean you require a greater centripetal force. What is providing this

 

[Doc Al]

doesn't going faster mean you require a greater centripetal force. What is providing this

I assume you're talking about the inside the loop version. In that case, the reaction force of the track provides additional force (in addition to the weight of the coaster) that creates the needed net force for the centripetal acceleration. The faster you go, the greater the contact force from the track on the coaster. (Up to the limit of the amount of force the track can provide.)

 

[jsmith613]

oh so as we go faster the contact force provided is greater and therefore the force required is met.
This will STOP when the track goes beyond its limits.
And also, from your animation, I presume my assumption that on the inside of the loop (that the car does not touch the track at the top) is wrong.

Does the reaction force thing apply to above loop roller coasters as well

 

[Doc Al]

oh so as we go faster the contact force provided is greater and therefore the force required is met.

Right.

This will STOP when the track goes beyond its limits.

The track will start to fall apart.

And also, from your animation, I presume my assumption that on the inside of the loop (that the car does not touch the track at the top) is wrong.

If you are going at the right speed, then at the top the contact force will be zero.

Does the reaction force thing apply to above loop roller coasters as well

Sure. But since the contact force acts upward at the top (instead of downward) the 'answer' will be different. For above the loop coasters there's a maximum speed to maintain contact at the top.

 

[jsmith613]

If you are going at the right speed, then at the top the contact force will be zero.

what speed is this

 

[BruceW]

$\frac{v^2}{r}=g$ Then the contact force is zero, since gravity provides the required centripetal force.

 

[jsmith613]

so the maximum speed on a on-top loop is v = sqrt(gr)
and for a underneath loop that is the mimnimum speed.?

 

[Doc Al]

so the maximum speed on a loop is v = sqrt(gr)
and for a non-loop that is the mimnimum speed.?

When going inside a loop, that's the minimum speed at the top to maintain contact; when going over a loop, it's the maximum speed.

 

[jsmith613]

ye, right.
ok thanks so much :)

 

[jsmith613]

I know this post was rather old but I was revising this topic and just wanted to re-clarify things:

we discussed before the minimium speed at the top of a vertical circle is sqrt(gr) and at the bottom it is sqrt(5gr).

Let us assume we have a net force towards the centre (mv^2/r + Tension) greater than mv^2/r
How can the object move in a circle.
I though mv^2/r was the EXACT force required to move in a circle.
OR is this the MINIMUM force required. If it is the MINIMUM force things make more sense

 

[Doc Al]

I know this post was rather old but I was revising this topic and just wanted to re-clarify things:

we discussed before the minimium speed at the top of a vertical circle is sqrt(gr) and at the bottom it is sqrt(5gr).

OK. To maintain contact on the inside of a vertical loop, you need a minimum speed of sqrt(gr) at the top; that translates to a minimum speed of sqrt(5gr) at the bottom to make it to the the top with sufficient speed.

Let us assume we have a net force towards the centre (mv^2/r + Tension) greater than mv^2/r
How can the object move in a circle.
I though mv^2/r was the EXACT force required to move in a circle.
OR is this the MINIMUM force required. If it is the MINIMUM force things make more sense

mv^2/r is the net force towards the center on something that is executing circular motion. The forces acting on the body will be gravity and the normal force.

 

[jsmith613]

ok so if 20 kg moves at a speed of 10 m/s and a radius of 3m then the NET force i need (Fc) is 666.666N

So normally the reaction force should accommodate this? right? but then if moving on the inside of the circle we can say that mg + reaction force = mv^2/r so the mv^2/r would only be the same if the reaction force provides all the rest of the force.
what if the mg + reaction force > 666.666N what would happen then?

 

[vela]

It works both ways. If you know the speed and the radius, you can find the acceleration. If you have the acceleration and the speed, you can find the radius.

If F_c>667 N and the 20-kg object is moving at 10 m/s, the radius of curvature of its path will be smaller than 3 m.

 

[jsmith613]

It works both ways. If you know the speed and the radius, you can find the acceleration. If you have the acceleration and the speed, you can find the radius.

If F_c>667 N and the 20-kg object is moving at 10 m/s, the radius of curvature of its path will be smaller than 3 m.

yes but let's imagine it was on a track. would it just fall off the track?