# Tension in the thread: Centripetal Force in the Vertical plane

1. Jun 2, 2014

### mistermill

I get the theory:

Fc = Ft + Fg

Uniform circular motion, in the vertical plane.

Fc = net force = centripetal force
Ft = force of tension in the string
Fg = force of gravity

so at the top, with vector addition:

Ft = Fc- Fg

And if we define 'up' as positive, then should the equation look like

-Ft = -(mv^2/r) - (-mg)

because all these forces point down (to the centre)

or just

Ft = -(mv^2/r) - (-mg)

because Ft should COME OUT negative after the arithmetic?

Sometimes, in some questions, the answer key has Ft positive. Why? Does that mean the system is impossible (ie/ velocity is not sufficient) or just that the answer keys are wrong? My sense is that IF Ft comes out positive, that would imply that the string pushes outward, which is not what physics tells us. All the forces are directed toward the centre of the circle (down at the top of the circle). Period.

Should we just define "to the centre" as positive and then this won't be a problem? Help!?

Last edited: Jun 2, 2014
2. Jun 2, 2014

### PhanthomJay

It is usually always best to define the direction of the net force (centripetal in this case) as positive. This avoids confusion with the minus sign. Note also that tension forces always pull away from the objects on which they act. So at the top, F_c = mg + T = mv^2/r, that is
T =mv^2/r - mg.

At the bottom, T -mg = mv^2/r, or
T = mv^2/r + mg.

At the quarter circle point
T = mv^2/r.

Note that T always comes put positive. If it comes out negative, something is amiss.

3. Jun 2, 2014

### BvU

Hello Mill, and welcome to PF.

I get an approximate idea of what this is about, but you would do well to be more concrete and precise.

As you know, Newton's law is $\vec F = m \vec a$, with $\vec F = \sum \vec F_i$ if more forces are acting. Applied to the situation you describe, you get something like $m \vec a = \vec F_t + m\vec g$. $\vec F_t$ is the force the string exerts on the rotating mass, and $m \vec a$ is the resulting force that makes the mass go round in a circle. The magnitude is mv^2/r. No sign.

Completely independent of the choice of coordinate system to express these vectors in (e.g. what is up, what units, etc.) as it should be.

Next you want to fill in some numbers to do calculations, and you need signs, a direction, etc.

Would want to argue the case against choosing like he does, because that gets you a rotating frame of reference, which is better avoided (unless of course when absolutely necessary). Better to choose as you did: up is positive. At the top of the circular trajectory, that means $\vec g$ = -9.81 m/s2 and $\vec F_t$ is negative too. Together they provide - mv^2/r towards the center of rotation.

4. Jun 2, 2014

### PhanthomJay

I guess you're in love with the minus sign. It is the number one reason for errors in Intro Physics. I prefer F_net = ma rather than -F_net = -ma.

5. Jun 2, 2014

### mistermill

My text in its description, when it is teaching gives an example, where the ft comes out negative, and explains it away with : the direction is downward.

But in a question at the end of the chapter, the answer in the back has Ft as positive. (at the top of the vertical circle).

I checked that the v = √(gr) and sometimes, in some of the assigned questions, the v is just not sufficient, so I thought, well what a stupid question!

Here is an example that I had for homework, and I calculated the result, but in the answer key, the Ft at the top is positive:

7. A 2.5 kg mass is attached to the end of a 3.0 m long rope and spun in a vertical circle at a speed of 5.6 m/s. Determine the maximum and minimum tensions in the rope. (6 marks, 3 marks each) [max = 51 N, min = 1.6 N]

Fc = Ft + Fg

rearrange:

Ft = Fc - Fg

top:

-Ft = -mv^2/r - (-mg)

-Ft = -1.6 N

Ft = 1.6N, which is the given answer, but shouldn't it be negative, ie/ toward the centre of the circle?

Another way I have seen is that the explanation just says:

The tension at the top is its lowest therefore T = Fc - mg
the tension at the bottom is its highest therefore T = Fc + mg :

this line of thinking seems to get the right answer, but it seems mathematically cheap, and does it really only solve for the magnitude?

6. Jun 2, 2014

### PhanthomJay

I cautioned that the pesky minus sign can bury you into a hopeless abyss. Sometimes it means you assumed the wrong direction for the force. But if you always draw your tension forces pulling away from the object, then it's direction is correct, and the minus sign if you get one for tension means something else is wrong, like for example if the speed at the top is less than sq rt of rg, your tension comes out negative meaning the object cannot whirl in a vertical circle. It takes a lot of practice to conquer the minus sign. Think positive!

7. Jun 2, 2014

### mistermill

So are the free body diagrams in my text incorrect? The tension at the top of the circle SHOULD be UP? It should be positive??

It should always be:

The tension at the top T = Fcentripetal - Fgravity and the result is positive.

F centripetal (net force) is down, Fgravity is down, tension is up.

(provided there is sufficient velocity, of course.)

So any negative result for Tension at the top, is probably an error? Or an over-simplification that is physically incorrect?

8. Jun 2, 2014

### mattt

One cute question for mistermill:

What is the net vertical force at the quarter circle point ( we know the net horizontal force at that point is mv^2/r ) ?

:-)

9. Jun 2, 2014

### mistermill

I think the net vertical force = 0 at that point. mg = T Why is that cute? Does it matter if the mass is going up or coming down?

I wonder how to 'teach' this concept of the tension being Fc - mg at the top and Fc + mg at the bottom, without introducing the signs, because the net force is always to the centre, but the direction of Ft is also. My text shows Ft toward the centre in the free body diagram.

10. Jun 2, 2014

### mattt

The net vertical force at that point is zero, because in a uniform circular motion the net tangential force (at any point of the trajectory) is always zero ( why? ), and at the quarter circle point vertical = tangential.

At that point, $$\vec{T}$$ and $$m\vec{g}$$ are orthogonal (perpendicular).

At that point, $$\vec{T}$$ is the net normal force (that is causing the normal=central acceleration), so....there must be another force (vertical and upwards at that point) that compensates the weight force.

In other words, if there is only $$m\vec{g}$$ and $$\vec{T}$$ you could never achieve a uniform circular motion in a vertical plane.

11. Jun 2, 2014

### mistermill

So the intro physics that we are taught is basically impossible.

Does everyone agree?

All you people that are smarter than me, please chime in!

12. Jun 2, 2014

### PhanthomJay

what do you mean it should be up?? I will say it again for the third time: Tension forces always pull AWAY from the objects on which they act. Put the tail of the tension force vector on the object, and the arrow of the tension force is well at the other end, directed along the rope. I don't care if the tension force is vertical or horizontal or at 45 degrees or at any direction....it pulls on the object, pointing away from it, not toward it.
yes, but do not think of centripetal force as a unique force of its own. The centripetal force is just the net force toward the center of the circle. It is made up of all forces acting radially toward or away from the center of the circle. It is newtons 2nd law in play here, in the centripetal direction.
yes
yes
What?? The tension acts down

13. Jun 3, 2014

### mistermill

When you said 'away' I got startled and thought you meant away from the centre.

Whew!

So the FBD are right. And if I use

Fc = Fg + Ft for the top and rearrange

Ft = Fc - Fg

And put in the signs

-Ft = -Fc - -Fg

and it comes out negative, then the question is physically impossible. Or I can leave the

Ft = Fc + Fg and just say Ft acts toward the centre of the circle.

Got it. Thanks!!

Now what mattt was saying makes sense: there is something else there, but we don't get to learn about it yet. I think it is pretty weird that all (gravity and tension) the forces act down, but nothing 'holds' the mass on the string UP. I just chalked it up to the constantly changing direction of the velocity.

Super weird.

14. Jun 3, 2014

### mistermill

Does the string exert a 'normal' force? Why do we not account for it?

15. Jun 3, 2014

### mattt

With "normal" I meant "normal to the trajectory" (perpendicular to the trajectory).

In a uniform circular motion, the total force (net force) is always (at any point) central = radial (directed to the center of the circle=trajectory), i.e. normal to the trajectory.

The string always exerts a force that is directed along the string and so is always normal (perpendicular) to the trajectory of the body in this uniform circular motion situation. It is what I called $$\vec{T}$$.

Do you understand why with only $$\vec{T}$$ and $$m\vec{g}$$ you could never achieve a uniform circular motion in a vertical plane?

Have you studied the "acceleration vector = tangential acceleration vector + normal acceleration vector"?

In a uniform circular motion, the tangential acceleration vector is zero at all points of the trajectory (the circle). In a uniform circular motion in a vertical plane, the weight force has a non-zero tangential (tangential to the trajectory) component in all points (but two, the top point and the bottom point of the trajectory=circle), so there must be another force to compensate the "non-zero tangential component of the weight force" at all those other points of the circle (all points but the top and the botton points) if you really have a uniform circular motion in a vertical plane.

16. Jun 3, 2014

### PhanthomJay

You can have a Ferris wheel move in a vertical circle in uniform circular motion. There are forces that counteract gravity at say the quarter point . But in the case of the object with a cord whirling in a vertical circle, you do not in the ordinary case have uniform circular motion. Such motion is non uniform with changing speeds throughout. At the quarter circle, only gravity is acting down, so therefore there is tangential acceleration acting down, the object is thus accelerating down at that point at a rate equal to g. And accelerating radially at v^2/r from the inwardly directed horizontal tension force.

17. Jun 4, 2014

### BvU

I am afraid we are now discussing over Mills head: his exercise does have a rope spinning and it does state one and only one speed. Distractingly unphysical, but Mills problem is in a different ball park: tension in the rope.

For Mills: grab both ends of a rope and pull. What you feel is the tension the rope transfers from one end to the other. Draw a force diagram with arrows: In the rope, on the rope, at both ends. Four arrows, all equally big (I should hope).
You could also draw arrows at intermediate points, again equally big and pointing opposite.
In physics, tension in a rope is 0 or positive. No meaningful physics to be done pushing the ends inwards.
No meaningful physics to be done exercising forces sideways either: A rope (spring, rubber band) can only transfer force longitudinally.

18. Jun 4, 2014

### mistermill

I like it when things are over my head. I appreciate the insight. I have read ahead to tangential speeds. And I understand there is a difference in a vertical circle with a rigid object and a vertical circle with a string.

I just find it weird that my text points the Tension down at the top (to the centre) and sometimes call that positive tension, and in some answers calls it negative. and when it comes out negative, it gets explained aways as "to the centre" but then when it comes out positive, the books says "and that is to the centre"

I want the WHY, not just the right answer.

:)

19. Jun 4, 2014

### PhanthomJay

So what happens at the point when the object has rotated 45 degrees from the top? The tension force acting on the object pulls away from the object acting inward toward the center of the circle at a 45 degree angle down from the horizontal. Is the tension a plus or a minus? Hmmm???

You've got to go over problems over and over again until you conquer the minus sign. It is NOT an easy thing to do!

Let's forget this circular motion stuff and if you don't mind, I would like for you to consider a string attached to the ceiling, you know, an ideal massless inextensible string attached to a ceiling. Now pull down on that rope with a force of 50 N. I hope you understand that the tension in the rope is 50 N, and it points up away from your hand toward the ceiling...pointing up .....but if you look at the ceiling, it points down from the ceiling toward your hand. This is newton's third law. In any case, the tension in the string is 50 N. No plus or minus sign. Just 50 N.

Now imagine the string is attached to the floor. Now pull up on that string with a force of 50 N. The tension force pulls away from your hand acting toward the floor...pointing down ..but if you look at the floor, it points up from the floor toward your hand. The tension in the string is still 50 N. Period. It's easy, no?