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.)jsmith613 said: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
I think you mean on the outside. (On the inside of the loop is different, since the 'reaction' force points in a different direction.)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).
Good!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)
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.(this contradicts the formula I found which is wierd)
jsmith613 said: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 said:I think you mean on the outside. (On the inside of the loop is different, since the 'reaction' force points in a different direction.)
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.
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.)jsmith613 said:first are all your responses based on the roller coaster being on the inside or outside? (i.e: "Good")
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.)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
Again, you seem to be mixed up between outside and inside the loop. Here's an illustration of the coaster being inside the loop: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?
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 said:doesn't going faster mean you require a greater centripetal force. What is providing this
Right.jsmith613 said:oh so as we go faster the contact force provided is greater and therefore the force required is met.
The track will start to fall apart.This will STOP when the track goes beyond its limits.
If you are going at the right speed, then at the top the contact force will be zero.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.
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.Does the reaction force thing apply to above loop roller coasters as well
Doc Al said:If you are going at the right speed, then at the top the contact force will be zero.
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 said:so the maximum speed on a loop is v = sqrt(gr)
and for a non-loop that is the mimnimum speed.?
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.jsmith613 said: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).
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.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
vela said: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 Fc>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.
I think you understand this, but just to reiterate BruceW's point, the normal force would never actually cause the roller coaster to come off the track. It would just be enough to keep roller coaster following the circular path. It's just like the normal force exerted on a book by a table is just enough to keep the book from falling downward. It would never be so large as to cause the book to accelerate upward and jump off the table. In other words, your hypothetical situation that mg+reaction > mv2/r (with a non-zero reaction force) is physically impossible.jsmith613 said: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?
Doc Al said: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.
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.
Right. There's no way to travel on the outside bottom of a loop without something to pull you up. Normal force and gravity won't help you.jsmith613 said:I know this discussion was along time ago but I want to check something
If I am moving around the OUTSIDE OF a circle, surely at the bottom of the circle (UNLESS I have a second pair of wheels to give me a reaction force big enough to stay on the track) I HAVE TO FALL OFF (regardless of my speed)
Is this right?
Doc Al said:Right. There's no way to travel on the outside bottom of a loop without something to pull you up. Normal force and gravity won't help you.
Right.jsmith613 said:If the object moved at 2 m/s then the centripetal force required is 0.4N
Which is way too much!the force provided at the top is 10N
Don't think in terms of minimum centripetal force, but minimum normal (reaction) force. If the reaction force goes below zero, that means you are losing contact. (And the reaction force can only be positive.)surely this means the object should STAY in the circle as the minimum force is provided. But this seems to contradict the rule min speed is sqrt(gr)
where have I gone wrong??
Doc Al said:Right.
Which is way too much!
Don't think in terms of minimum centripetal force, but minimum normal (reaction) force. If the reaction force goes below zero, that means you are losing contact. (And the reaction force can only be positive.)
The minimum speed is what you need to give you just exactly zero reaction force at the top; no problem if you go faster, but you'll fall off the loop if you go slower.
Doc Al said:Yes.
Another way to understand what's going on when you're inside the circle at the top is to realize that the minimum force on you will equal your weight. So any speed which requires less than that amount of centripetal force cannot keep you on the track. The excessive force of your weight will pull you off the track and into the air.
jsmith613 said:The most terrible thing about this topic is every time I revise it or revist this discussion I always get confused!
With the exam coming up I am now revisiting this topic.
Object moving around the inside of the track
So we discussed earlier that mv^2/r (centripetal force) is the net force towards the center on something that is executing circular motion
This force needs to be provided by associated forces such as reaction force or weight
At the top of the circle, if the radius is 10m and the speed is 30 m/s then the centripetal force required to keep the object (of mass 1kg) in a circle is 90N
the weight of the object is 10N so I would presume the reaction force provides the other 80N.
If the object moved at 2 m/s then the centripetal force required is 0.4N
the force provided at the top is 10N
surely this means the object should STAY in the circle as the minimum force is provided. But this seems to contradict the rule min speed is sqrt(gr)
where have I gone wrong??
thanks
PeterO said:So near and yet so far!
eg: a lift accelerates up at 2m/s/s - how heavy does the 100kg occupant feel? [use g=10 for simplicity]
Net Force [red arrow] 200N UP
Weight Force [black arrow] 1000N DOWN.
Green arrow is thus 1200N up
1200N up is how strongly a floor normally pushes on a 120kg person, so the occupant's apparent weight is 120kg.
Indeed the colour coding is useful in many problems: get your self three different pens and use them always.
Often people forget the weight force in a problem, but given that the first thing you always draw is the black arrow, there can be no omissions.
jsmith613 said:Surely green = 800N
Weight = Acceleration + Reaction force
R = W - A
R = 1000 - 200
R = 800N
so the reaction force will be 800N
PeterO said:NO!
When a lift accelerates up - ie it first moves off when you are traveling up - do you feel heavier or lighter?
The weight force is down - black arrow
The net force is up - the Red arrow
The Green arrow goes from the bottom of the Black arrow, all the way up to the top of the black arrow, then even further to the top of the Red arrow.
As for "Weight = Acceleration + Reaction force" I was simple using
Apparent Weight = Reaction Force
Perhaps your "Weight - acceleration" should be accounting for the fact that acceleration is in one direction, while gravity is in the other, so one of them should be negative
So 1000 - -200 = 1200 ?
When I just use the numbers, I sometimes go wrong.
When I use the Arrows, and look at what is going where, I always get the right answer.
The net force is always in the direction of the acceleration, not necessarily in the direction of the velocity. (In this example, the acceleration and velocity of the lift are both up.)jsmith613 said:why is the net force up? the net force is not always in the direction of motion??
