Understanding centripetal force and free body diagram

[vela]

Yes. The object's path will curve more sharply than the track, so its path and the track will diverge, i.e., it falls off.

 

[jsmith613]

awesome
thanks

 

[BruceW]

Yeah, the idea with the tension force is that it is only as much as is required to stop the car from breaking through the track. So the tension force will never cause the car to fall off the track toward the centre of the circle.

The only way the car falls off the track is if it is not going fast enough around the track, so that gravity on its own pulls the car off the track (in this case, the tension is zero).

 

[vela]

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?

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.

I thought what you were really asking is what happens if F_net, whatever forces it includes, is greater than mv²/r. In that case, the roller coaster would come off the track. It would also leave the circular track if F_net < mv²/r (probably breaking the track). It's only when F_net = mv²/r does it stay on the track.

 

[jsmith613]

thanks for this!

 

[BruceW]

what a polite young man :)

 

[jsmith613]

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.

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]

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?

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]

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.

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

 

[Doc Al]

If the object moved at 2 m/s then the centripetal force required is 0.4N

Right.

the force provided at the top is 10N

Which is way too much!

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??

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.

 

[jsmith613]

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.

well I guess if your inside the circle, you go faster the track provides the extra reaction force

so your suggesting the best way to do this is to calculate the reaction for needed to provide the centripetal force
in this case
0.4 = 10 + R
R = -9.6
as R < 0 this cannot be so it falls off

If we had a centripetal force of 300 then
300 = 10 + R
R = 290
R > 0 so carries on in circle (if there is the capability for the track to provide this extra force?)

 

[Doc Al]

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]

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.

thanks

would it be possible to look at this thread:
https://www.physicsforums.com/showthread.php?p=3838852&posted=1#post3838852

It is a similar question to centripetal force but with cars moving around a hoizontal track and frictional forces providing centripetal acceleration

 

[PeterO]

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

So near and yet so far!

I will explain it two ways:
Test by extreme:- Why so fast? Let the speed drop to 0m/s, then the "centripetal Force" required is zero, and we are providing 10N. Will it fall off ?

Well if it is on the inside of the track - YES. If it is on the outside of the track - NO.

Colour codes to keep track:

I recommend to my students to use 3 [or more] coloured pens during every circular motion problem - and use them to draw vectors of the forced involved.

The most popular code is:
Weight - Black
Resultant force - Red
Reaction force - Green

Having drawn our object on the circular track - you then draw in the vectors you know.
[The scale is arbitrary - but an arrow representing 0.4 is a lot shorter than an arrow representing 10]

a Black, vertical arrow pointing down - gravity, so the weight Force is always down - of length 10

A Red result and arrow pointing towards the centre - in this case down - of length 0.4

The Green arrow representing the force the environment must provide is then drawn from the head of the black arrow, to the head of the Red arrow.
This is actually showing that Weight + "Environment" = net Force.

In this case the green vector is 9.6N UP. can the track [or anything else touching the object] provide a force of 9.6N UP.

On the inside of the track - NO, so the situation won't work; meaning the object will NOT follow the track - it will fall off.
On the outside of the track - YES, the reaction force will be fine.

Now consider the firs case.
The Black arrow is still 10 units down, The Red arrow is 90 units down [ie bigger], so the green arrow is also down [and 80 units long]

Can the track supply a downward force.
On the inside - YES; it follows the track
On the outside - NO the object would fly off the track.

The same colour codes can be used for people in lifts as the lift starts up/down or stops from up/down.

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]

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.

Surely green = 800N
Weight = Acceleration + Reaction force
R = W - A
R = 1000 - 200
R = 800N
so the reaction force will be 800N

 

[PeterO]

Surely green = 800N
Weight = Acceleration + Reaction force
R = W - A
R = 1000 - 200
R = 800N
so the reaction force will be 800N

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.

 

[jsmith613]

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.

why is the net force up? the net force is not always in the direction of motion??

 

[Doc Al]

why is the net force up? the net force is not always in the direction of motion??

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]

of course...thanks