Tension in a pendulum

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evoviiigsr

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I know I should know this, but remind me, how does the magnitude of tension change in a pendulum oscillating in SHM? Is it at its max in the resting position? If so, by what factor is it decreased at any other points? Thanks
 
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evoviiigsr said:
I know I should know this, but remind me, how does the magnitude of tension change in a pendulum oscillating in SHM? Is it at its max in the resting position? If so, by what factor is it decreased at any other points? Thanks
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The tension in a pendulum is given by this formula:

T = mgcos(theta) + Mv^2/r

So when the pendulum is at the very bottom, not only does it have mg but also centripedal force. The tension is hence greatest at the bottom and least when it's at its highest point.

Edit: draw a FBD and you'll see what I mean.
 
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So, when its at the lowest point, T=mgcos(theta) + mv^2/ r.. got that

The centripetal force doesn't act only at the lowest point, but throughout the swing, correct?

At the highest point, where theta is high (first term is low), is there any centripetal force? Since velocity is momentarily zero as direction changes, that would suggest no. But intuitively it doesn't make sense for the centripetal force to disappear and then reappear on the swing back down.
 
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evoviiigsr said:
So, when its at the lowest point, T=mgcos(theta) + mv^2/ r.. got that

The centripetal force doesn't act only at the lowest point, but throughout the swing, correct?

At the highest point, where theta is high (first term is low), is there any centripetal force? Since velocity is momentarily zero as direction changes, that would suggest no. But intuitively it doesn't make sense for the centripetal force to disappear and then reappear on the swing back down.
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There's this common misconception that centripetal force is an actual force; it actually isn't. Centripetal force is the net force that points inwards towards the center that keeps the mass in motion. So for instance, if the theta is at 90 degrees, cosine of 90 is 0 which means that at this point in time, the pendulum is pretty much free falling and there is not centripetal force towards the center. If you can imagine the pendulum swinging full 360, the centripetal force is greatest when the pendulum is right above the rotating axis, because at this point mg is pointing directly downward towards the center.

To sum it all up:

Centripetal force is greatest while the pendulum is right above the rotating axis, and at 0 when it's 90 degrees on either side and at the bottom.

The tension is greatest at the bottom and smallest at the top.


Edit: I have a correction to make. I'm actually quite confused right now myself. According to wiki, The centripetal force is the external force required to make a body follow a circular path at constant speed. The force is directed inward, toward the center of the circle.

In this case, I would have to say that it's actually the tension of the rope, that's the centripetal force by definition. The thing about this problem is that you can't really use centripetal force because the velocity is not constant, you have a mass that's accelerated downwards by gravity. And after this I just realized that in this problem; centripetal force is DEFINITELY the same as T. Because by the definition above, anything that keeps the object in constant velocity circling is the centripetal force. So there you have it. CENTRIPETAL = Tension.
 
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Hmm interesting..

you guys seen a problem like this on the MCAT.. relating tension and pendulums.. I've really only seen period/mass ones
 
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If you draw a triangle below the pendulum bob with mg as the hypotneuse (straight down) and resolve the vectors, the tension at any point in the swing is equal to MGcos(theta).

T = MGCOS(THETA) so if the mass is hanging straight down, the tension is just mg. If the mass is horizontal, then its 0 since its in free fall.

The centripetal force is not a force itself. It's a contribution by a variety of forces including, electrostatic, contact forces, gravitational. The centripetal force of a car on a flat plane is just friction. The centripetal force of a car on a banked plane is gravity + friction. The centripetal force of a moon orbiting a planet is due to gravitational forces.

😀 I think haha
 
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vicinihil said:
If you draw a triangle below the pendulum bob with mg as the hypotneuse (straight down) and resolve the vectors, the tension at any point in the swing is equal to MGcos(theta).

T = MGCOS(THETA) so if the mass is hanging straight down, the tension is just mg. If the mass is horizontal, then its 0 since its in free fall.

The centripetal force is not a force itself. It's a contribution by a variety of forces including, electrostatic, contact forces, gravitational. The centripetal force of a car on a flat plane is just friction. The centripetal force of a car on a banked plane is gravity + friction. The centripetal force of a moon orbiting a planet is due to gravitational forces.

😀 I think haha
Click to expand...

At the bottom of the string the Tension is Mg + MV^2/r, not just MG.
 
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vicinihil said:
you sure about that? hmm I think that makes sense, but I thought Fc isn't a real force? so where is it coming from.
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Ok, so after doing some serious research this is how it goes. Like in my past 2 post above I believe, Fc is the net force that points inwards towards the center, mV^2/r. At the bottom of the string, T is point towards the center and mg is pointing downwards. So the eqn goes: T- mg = mv^2/r, solve for T and you get T = mv^2/r + mg.

Although centripetal force is not a "real" force. Centrifugal force is, which is the reaction pair of centripetal force (Newton's 3rd law). It points downwards; so if you solve it another way, you have T point up, mg down, and centrifugal force which also points down. You set them equal to each other because there is no net acceleration in the vertical direction which makes it T = mg + mv^2/r. Either way, you get the same answer.
 
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EECStoMed said:
Ok, so after doing some serious research this is how it goes. Like in my past 2 post above I believe, Fc is the net force that points inwards towards the center, mV^2/r. At the bottom of the string, T is point towards the center and mg is pointing downwards. So the eqn goes: T- mg = mv^2/r, solve for T and you get T = mv^2/r + mg.

Although centripetal force is not a "real" force. Centrifugal force is, which is the reaction pair of centripetal force (Newton's 3rd law). It points downwards; so if you solve it another way, you have T point up, mg down, and centrifugal force which also points down. You set them equal to each other because there is no net acceleration in the vertical direction which makes it T = mg + mv^2/r. Either way, you get the same answer.
Click to expand...

whoa whoa whoa centrifugal force is definitely NOT a force if there's anything I learned. It's an imaginary force.

However, I like the T-mg = Fc which then becomes T = mv^2/r + mg

so what if a pendulum was at its apex of swing? would it be T = Fc since mgcos(theta) is 0?
 
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Oh boy, this is a great discussion. In reality, you are both right. Centrifugal force is both a real and a fictitious force. Strange, huh? Let me explain.

When you're in a car that is going around a corner, you feel a force pull you to the outside. This force is often called the centrifugal force, but it's not real. What you're really experiencing is not a centrifugal force but merely your inertia which is tangential to the rotation and therefore pushes you against the car as it turns. This comes about due to the fact that this situation involves a rotating reference frame.

However, the force of your body pressing against the side of the car as you turn is in fact a centrifugal force and it is a REAL force. Like was stated earlier, this is a reaction force which directly counters the centripetal force of the car pushing in on you.

Complex stuff.
 
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What's all this nonsense about centripetal force not being real. Common sense tell you that if it's not a real force then bodies/objects, in the absence of all other forces, would only travel only in a straight line as stated by Newton's first law. There's a centripetal force when a body moves in a (partially) circular path because there's a radial component of acceleration that always acts perpendicular to the velocity at every point of the (partially) circular path. The consequence is that the radial acceleration changes the direction of the velocity such that the time rate change of velocity vectorally points to the "center".
 
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Foghorn said:
What's all this nonsense about centripetal force not being real. Common sense tell you that if it's not a real force then bodies/objects, in the absence of all other forces, would only travel only in a straight line as stated by Newton's first law. There's a centripetal force when a body moves in a (partially) circular path because there's a radial component of acceleration that always acts perpendicular to the velocity at every point of the (partially) circular path. The consequence is that the radial acceleration changes the direction of the velocity such that the time rate change of velocity vectorally points to the "center".
Click to expand...

You're right, there's no question about the existence of centripetal force. The issue is in reference to its partner, centrifugal force.
 
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saqrfaraj said:
Oh boy, this is a great discussion. In reality, you are both right. Centrifugal force is both a real and a fictitious force. Strange, huh? Let me explain.

When you're in a car that is going around a corner, you feel a force pull you to the outside. This force is often called the centrifugal force, but it's not real. What you're really experiencing is not a centrifugal force but merely your inertia which is tangential to the rotation and therefore pushes you against the car as it turns. This comes about due to the fact that this situation involves a rotating reference frame.

However, the force of your body pressing against the side of the car as you turn is in fact a centrifugal force and it is a REAL force. Like was stated earlier, this is a reaction force which directly counters the centripetal force of the car pushing in on you.

Complex stuff.
Click to expand...

Inertia cannot be tangential to rotation. Inertia is not a vector. It has no direction.
 
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Foghorn said:
What's all this nonsense about centripetal force not being real. Common sense tell you that if it's not a real force then bodies/objects, in the absence of all other forces, would only travel only in a straight line as stated by Newton's first law. There's a centripetal force when a body moves in a (partially) circular path because there's a radial component of acceleration that always acts perpendicular to the velocity at every point of the (partially) circular path. The consequence is that the radial acceleration changes the direction of the velocity such that the time rate change of velocity vectorally points to the "center".
Click to expand...

Yeah I've always thought that it was a force like any other, like gravity for instance. And all of this talk about centrifugal force? Haven't really heard of it other than it being mentioned, doubt that we need to know it for the MCAT.
 
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I just wanted to say this: I HATE THIS **** RIGHT NOW. I've been scoring in the 13s and 14s for PS, yet after doing the EK 1001 series for Physics and G-chem I am FOREVER CONFUSING MYSELF MORE.

Can someone of authority, please tell me what exactly is centripetal force and settle this discussion once and for all?
 
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BrokenGlass said:
Inertia cannot be tangential to rotation. Inertia is not a vector. It has no direction.
Click to expand...

This is true but irrelevent. Due to inertia your body wants to maintain its direction of travel which is in the direction tangential to rotation. If you want to split hairs, being "tangential" alone does not imply a vector quantity because no direction is specified.
 
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EECStoMed said:
I just wanted to say this: I HATE THIS **** RIGHT NOW. I've been scoring in the 13s and 14s for PS, yet after doing the EK 1001 series for Physics and G-chem I am FOREVER CONFUSING MYSELF MORE.

Can someone of authority, please tell me what exactly is centripetal force and settle this discussion once and for all?
Click to expand...

I wouldn't say I'm "someone of authority," but I'll answer anyway. Centripetal force is merely a force required to make an object travel in a circle. The force is perpendicular to the direction of motion and points towards the center of the circle. Any force can be a centripetal force. Gravity, tension, etc.
 
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saqrfaraj said:
This is true but irrelevent. Due to inertia your body wants to maintain its direction of travel which is in the direction tangential to rotation. If you want to split hairs, being "tangential" alone does not imply a vector quantity because no direction is specified.
Click to expand...

It is both true and relevant. When someone reads your post and "learns" incorrect information from it, they will miss a question on the MCAT. Which I why I pointed this out. I wasn't trying to split hairs.

Also, I disagree about your interpretation of tangential. It does imply direction even if we don't know what the exact direction is.
 
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EECStoMed said:
I just wanted to say this: I HATE THIS **** RIGHT NOW. I've been scoring in the 13s and 14s for PS, yet after doing the EK 1001 series for Physics and G-chem I am FOREVER CONFUSING MYSELF MORE.

Can someone of authority, please tell me what exactly is centripetal force and settle this discussion once and for all?
Click to expand...

Centripetal force is the NET force which makes an object undergo uniform circular motion.
 
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BrokenGlass said:
It is both true and relevant. When someone reads your post and "learns" incorrect information from it, they will miss a question on the MCAT. Which I why I pointed this out. I wasn't trying to split hairs.

Also, I disagree about your interpretation of tangential. It does imply direction even if we don't know what the exact direction is.
Click to expand...

Well I definitely don't intend to mislead. However, I didn't say inertia was a force and I didn't say it was a vector.

Tangential implies two possible directions and so though it narrows down the possibilities it cannot fully describe a vector.
 
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saqrfaraj said:
Well I definitely don't intend to mislead. However, I didn't say inertia was a force and I didn't say it was a vector.

Tangential implies two possible directions and so though it narrows down the possibilities it cannot fully describe a vector.
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You didn't have to. Saying that inertia has direction automatically implies it's a vector. Tangential doesn't imply 2 possible directions. Tangential implies many more directions than that. But just because the precise direction is not known, it's incorrect to claim that the direction doesn't exist. Tangetnal "fully" describes a vector because there is magnitude and direction.
 
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BrokenGlass said:
You didn't have to. Saying that inertia has direction automatically implies it's a vector. Tangential doesn't imply 2 possible directions. Tangential implies many more directions than that. But just because the precise direction is not known, it's incorrect to claim that the direction doesn't exist. Tangetnal "fully" describes a vector because there is magnitude and direction.
Click to expand...

Sure, we could talk about tangential planes or whatnot, but that's hardly applicable to a pendulum problem. Regarding the issue of the vector, like you said, you need a magnitude and a direction. Unfortunately, a magnitude and two directions or a magnitude and 300 directions, will not suffice.

Let's think of a day-to-day example. How helpful is it if you ask directions to the store and someone responds "go 5 miles north and south"? Not very useful, huh?
 
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Either way, BrokenGlass, this is a pretty pointless argument/discussion we're having. I completely understand what you're saying, and I'm sure you see where I'm coming from too. Getting back to where this began, yes, I know inertia doesn't have a direction. I didn't think I implied it did, you thought otherwise, oh well.

Regarding tangents, I still don't think being tangent is enough to imply a definite direction. Universally, indicating something is tangent only narrows down the possible directions from infinite to slightly less infinite. Even when confining yourself to a plane there are still two possible directions. Yet if we were given a diagram of rotational motion I'm sure we'd both draw the same tangent vector, so why argue? 😀

So yep, that's just a long way of saying I'm going to bed.
 
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saqrfaraj said:
Sure, we could talk about tangential planes or whatnot, but that's hardly applicable to a pendulum problem. Regarding the issue of the vector, like you said, you need a magnitude and a direction. Unfortunately, a magnitude and two directions or a magnitude and 300 directions, will not suffice.

Let's think of a day-to-day example. How helpful is it if you ask directions to the store and someone responds "go 5 miles north and south"? Not very useful, huh?
Click to expand...

Now you are going on wild tangents (no pun intended). You are the one who mentioned planes. In fact, I don't even know where that came from. By many directions I meant the bob could be at one of many different positions as far as the arc is concerned, therefore there can be many different tangential vectors. You definitely misunderstood me there. If I tell you that there is force of tension pulling on a horse, are you gonna tell me that it's not a force because you don't know the exact direction?
 
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BrokenGlass said:
Now you are going on wild tangents (no pun intended). You are the one who mentioned planes. In fact, I don't even know where that came from. By many directions I meant the bob could be at one of many different positions as far as the arc is concerned, therefore there can be many different tangential vectors. You definitely misunderstood me there. If I tell you that there is force of tension pulling on a horse, are you gonna tell me that it's not a force because you don't know the exact direction?
Click to expand...

Yeah, I mentioned planes because you mentioned a tangent having the possibility of more than two directions. I though we were talking about a single point on an arc. When talking about a single point, you only have two directions a tangent could be in (we're talking 2D, right?) unless you're talking about a plane that is tangent at that point (3D). That's why I brought up planes. If you were talking about many points on the arc then that makes much more sense and planes don't have to enter into the discussion.

Regarding your horse example, I never meant that no vector existed. I ony meant that without a magnitude and a SINGLE direction you can't describe it definitively.
 
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saqrfaraj said:
You're right, there's no question about the existence of centripetal force. The issue is in reference to its partner, centrifugal force.
Click to expand...
That's nonsense too. Newton's 3rd law states that forces always act in pairs. Centrifugal force, aka inertia, is the counterpart to centripetal force.
 
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Foghorn said:
That's nonsense too. Newton's 3rd law states that forces always act in pairs. Centrifugal force, aka inertia, is the counterpart to centripetal force.
Click to expand...

You should really read my post which you initially quoted. The centrifugal force that supposedly "pulls" you to the side of the car as it turns a corner is a fictitious force. The definition of a fictitious force is an apparant force which comes about in a non-inertial reference frame. A rotating reference frame, a.k.a you in a car going in a circle, is a non-inertial reference frame. Therefore it does not exist.

You are refering to my second example. The force you put on the side of the car as you press against it is indeed a real force (as I said before). Like you said, it's the force that directly counters the centripetal force placed upon you by the car. No question there.

Centrifugal force is not, as you've said, the same as inertia. The fictitious centrifugal force that "pulls" you as the car rotates is inertia in disguise (because of the non-inertial reference frame), but inertia is not a force and cannot be the reaction force that counters the centripetal force.
 
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saqrfaraj said:
You should really read my post which you initially quoted. The centrifugal force that supposedly "pulls" you to the side of the car as it turns a corner is a fictitious force. The definition of a fictitious force is an apparant force which comes about in a non-inertial reference frame. A rotating reference frame, a.k.a you in a car going in a circle, is a non-inertial reference frame. Therefore it does not exist.

You are refering to my second example. The force you put on the side of the car as you press against it is indeed a real force (as I said before). Like you said, it's the force that directly counters the centripetal force placed upon you by the car. No question there.

Centrifugal force is not, as you've said, the same as inertia. The fictitious centrifugal force that "pulls" you as the car rotates is inertia in disguise (because of the non-inertial reference frame), but inertia is not a force and cannot be the reaction force that counters the centripetal force.
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what the hell is fictitious about Newton's 3rd law?

read noob, especially the 2nd to last paragraph of the quote below. xanthines' and shrike's explanations >>>> your babbling
http://forums.studentdoctor.net/showthread.php?p=5304821
Xanthines: Here is an explanation of centripetal force, using the example of your car making donuts in a level parking lot. In this case, the engine is providing the power for forward motion (by forward, I mean the direction of motion of the front of the car). The centripetal force equation Fc=(m)(v^2)/r describes the force required to keep the object in a circular path. The static friction force points towards the center of the circle, and this is the source of the centripetal force that keeps the car moving in its circular path for this example:

Fc = uFn = mv^2/r

Without a centripetal force, objects such as cars and yo-yos on strings will move in a direction that is tangential to the circular path they were on previously. Basically, what happens is that when you let go of the steering wheel, the car will move in a straight line that is tangential to the donuts you were making before. The car is not actually being pushed away from the center of the circle; rather, the centripetal force had been holding it in its circular motion, and now that centripetal force is gone. In addition, if the car moves fast enough, the force due to static friction will not be high enough to maintain circular motion. Again, that is because on level surfaces, the centripetal force is the friction force...I've heard the term "centripetal force requirement" used to describe this situation. This example shows us why it is that without friction, there can be no circular motion for cars. For example, when you try to make turns on icy patches of the road, you find that you cannot turn in a circle. There is no friction to provide a centripetal force, and thus you skid off the road into a ditch. Note that this is only true for level surfaces. On banked highways and race tracks, gravity comes into play.

To sum: centripetal force isn't really a new type of force, but rather is just the name used to indicate the net force pointing inwards for an object in circular motion. In the case of a car making donuts, the centripetal force is the static friction force. Since the static coefficient of friction is normally given as the maximum, you know that any more force applied to that object will overcome the friction, and the object will begin to slide in a direction tangent to the circle of motion (or in the case of the car, skid off the road.) Analogously, when a satellite moves fast enough, it will escape the earth's gravity:

G(Mearth)(Msatellite)/(r^2) is less than (msatellite)(v^2)/r

One final point: When most people use the term "centrifugal force," they are actually referring to what is commonly referred to by another physics term known as inertia.

Shrike: It's true that by "centrifugal force," non-physicists usually mean something else: either inertia, or centripetal force. But to clarify what Xanthines said, there *is* such a thing as centrifugal force. Recall that, by Newton's Third Law, for every force there's an equal and opposite force, and that the two constitute an action-reaction pair. Centrifugal force is the force that forms a pair with the centripetal force -- it's the force exerted by the body that's moving in a circle, on whatever's making it turn.

For example, in the case of a yo-yo whirled on a string, the force of the string on the yoyo is centripetal, and the force of the yo-yo on the string, and thus your finger, is centrifugal. In the case of the car driving in a circle, the ground exerts a centripetal foce on the tires, while the tires exert a centrifugal force on the ground. You might guess from these examples that it's centripetal force that we usually worry about, not centrifugal. You'd be right. Centrifugal force tends to be a little odd, and mostly irrelevant. I've never seen it matter on an MCAT problem.
Click to expand...

i agree with foghorn that all this discussion about whether centripetal and centrifugal forces are fictious being nonsense. you nubs are just arguing about semantics :laugh:
 
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heymanooh1 said:
what the hell is fictitious about Newton's 3rd law?

read noob, especially the 2nd to last paragraph of the quote below. xanthines' and shrike's explanations >>>> your babbling


i agree with foghorn that all this discussion about whether centripetal and centrifugal forces are fictious being nonsense. you nubs are just arguing about semantics :laugh:
Click to expand...

Now we're resorting to name calling? 😀 The info you posted is all great stuff and I agree with everything they say. Read what I've posted and you'll see.
 
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