Uniform circular motion with a pendulum

[Femme_physics]

Homework Statement

http://img685.imageshack.us/img685/9250/drawingmo.jpg

In the drawing is depicted a pendulum hammer of an impact device (in both projections). The hammer is made out of shaft AB which upon it is held pendulum OM. At point M of the pendulum is a hammer. According to experiment they settle the pendulum in a condition close to that in the drawing, release it and it drops freely, as depicted in the dashed line in the drawing. At the lowest point of the fall (point N), the hammer reaches the model and breaks it.

Given:

Hammer Weight = 200 [N]
Length of the pendulum = 800mm

A) Calculate the max tension in pendulum OM.

[there is also a clause B, but I'm getting stuck at A]

The Attempt at a Solution

I've written my question in the attempts

http://img824.imageshack.us/img824/3124/fy1q.jpg

http://img405.imageshack.us/img405/5020/fy2p.jpg

 

[tiny-tim]

Hi Femme_physics!

Sorry, I don't understand what you've done.

Try conservation of energy ​

 

[I like Serena]

Hi Fp :)

Short hint this time: it's not a uniform circular motion, so those formulas do not apply.

 

[Femme_physics]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

 

[Femme_physics]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

Thanks for letting me know!

 

[I like Serena]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

Thanks for letting me know!

Well, the hammer falls and accelerates while falling. That's why.

But you do have a formula that you've hardly used yet.
The one that says:

g h = v² / 2

That would apply...

 

[Femme_physics]

Well, the hammer falls and accelerates while falling. That's why.

But you do have a formula that you've hardly used yet.

How do you know I hardly used it? What else do you know about my usages?!?

g h = v² / 2

Hmm...how is this formula called?

We were given a list of such formulas but I don't see it. Or maybe I don't have yours and tiny-tim's super math vision feat.

http://img9.imageshack.us/img9/4261/listofsuchformulas.jpg

 

[tiny-tim]

We were given a list of such formulas but I don't see it.

It's the formula in the bottom-right box.

It's the conservation of energy equation for something falling vertically.

Since energy depends only on the endpoints, and not on how you get there (direct or devious),
the same conservation of energy equation applies to this circular motion.

 

[Femme_physics]

Thanks, tiny-tim! I wish he's given us NAMES to all the formulas, and not just the formulas themselves. I guess next class a decent question to the lecturer would be "can you name all those formulas?"

It's the conservation of energy equation for something falling vertically.

Duly noted and written.

Since energy depends only on the endpoints, and not on how you get there (direct or devious),

Oh yea, we were even told that in biology.

the same conservation of energy equation applies to this circular motion.

*nods*

Then I shall spur myself to action and reply back sometimes later! :) Thanks, tiny-tim.

 

[tiny-tim]

Thanks, tiny-tim! I wish he's given us NAMES to all the formulas, and not just the formulas themselves. I guess next class a decent question to the lecturer would be "can you name all those formulas?"

They don't have names.

This one is only conservation of energy if a happens to be gravity.

Oh yea, we were even told that in biology.

erm … it doesn't work in biology …

you use energy just standing still! ​

 

[I like Serena]

How do you know I hardly used it? What else do you know about my usages?!?

There!
https://www.physicsforums.com/showpost.php?p=3297647&postcount=24

My bad, I have to admit it looks a little different, and I guess it takes a bit of a super math/physics vision feat to see it.
You will, just give it another level, and then you will have this feat as well!

Hmm...how is this formula called?

Yeah, conservation of energy (for gravity and kinetic energy).

More specifically it is:

E = m g h + m v² / 2

This is the energy of the system that never changes (assuming no friction).
So after any change, the energy E before, must be equal to the energy after.

Perhaps it's premature for you to see this, but I think you'll get it soon enough and it probably doesn't hurt if you'd recognize it.

 

[Femme_physics]

You will, just give it another level, and then you will have this feat as well!

*looks at character sheet*

This says I need 2948xp to ding! You have any ideas how many tough physics bosses problems I'll have to solve to get there?!? :(
I need an item that imbues me with that power ;)

They don't have names.

This one is only conservation of energy if a happens to be gravity.

Really? Nameless formulas? Well, someone must've come up with them and named them sometimes in the past. It can't be that he just came up with it and left it nameless?

erm … it doesn't work in biology …

I think we may have studied the physical properties that are also important in biology, but I clearly and distinctly recall this quote

"Since energy depends only on the endpoints, and not on how you get there (direct or devious),"

Perhaps it's premature for you to see this, but I think you'll get it soon enough and it probably doesn't hurt if you'd recognize it.

I tend to only get stuff after lots of problem solving, it appears. That's also how I like it :)

I used the diameter for "h" -- is that right? it's the change in position, delta x. So, the change in position is 1.6m!

http://img546.imageshack.us/img546/9615/vsolutiona.jpg

 

[tiny-tim]

I used the diameter for "h" -- is that right? it's the change in position, delta x. So, the change in position is 1.6m!

you're assuming that α = 0

 

[Femme_physics]

you're assuming that α = 0

Oh yes, sorry I didn't mention that. A rather important fact!

We're told to:

"ignore the weight of the shaft and the pendulum and assume the angle at the beginning of the motion is 0."

 

[tiny-tim]

 

[Femme_physics]

Ah...I missed that didn't you? ;)

Now that I got V I'm a little bit hesitant how to continue.

Hi Fp :)

Short hint this time: it's not a uniform circular motion, so those formulas do not apply.

Even now? Because, to find what they ask of me (tension at pendulum) don't I need to find the normal force? And doesn't that mean sum of all forces?...

[edit: heading out to college to help a student in statics :) , might not reply for a while]

 

[I like Serena]

Ah...I missed that didn't you? ;)

Now that I got V I'm a little bit hesitant how to continue.



Even now? Because, to find what they ask of me (tension at pendulum) don't I need to find the normal force? And doesn't that mean sum of all forces?...

[edit: heading out to college to help a student in statics :) , might not reply for a while]

All right!

I'll give you a couple more hints...

What forces act (or are required to act) on the hammer?
And at what point of the trajectory will these forces generate the greatest tension in the pendulum?
Finally, how large will these forces be at that point?


EDIT: As for your previous calculations ...
I'm missing ...
A little more of you! :P

 

[tiny-tim]

Now that I got V I'm a little bit hesitant how to continue.

Find the centripetal acceleration …

then use net force = mass times acceleration. ​

 

[Femme_physics]

And at what point of the trajectory will these forces generate the greatest tension in the pendulum?

Clearly when the hammer drops and just hits the sample

Finally, how large will these forces be at that point?

Well, I'll need to find out T before I can tell you. So I could tell how long does it take to make half a circle.

Hmm...how do I find it T? Do I use this formula? (I hope I used it correctly)

http://img39.imageshack.us/img39/783/triedtodecent.jpg

EDIT: As for your previous calculations ...
I'm missing ...
A little more of you! :P

;) I would but I'd hate for people to start to think I'm modeling and not trying to do my homework lol

Find the centripetal acceleration …
then use net force = mass times acceleration.

Hmm...I'm still missing T to find a, (since "a" has "v" in its formula, and "v" has "t" in its formula. It all comes down to T, it appears!)

Of course, if my above calculation is correct, then I got T. Is that right?

 

[I like Serena]

T is the time for a complete revolution of a uniform circular motion.

It's a pity that the hammer just falls and breaks something and is not neatly swinging around at an even speed to break something else.

The formula you need is the one that relates the (required) centripetal force to speed and radius.

That is:

F = m v² / R

 

[Femme_physics]

It's a pity that the hammer just falls and breaks something and is not neatly swinging around at an even speed to break something else.

More destruction! More destruction! :D

The formula you need is the one that relates the (required) centripetal force to speed and radius.

Brilliant.


And we were given this list of formulas.

http://img24.imageshack.us/img24/6849/frfrh.jpg

I can see that formula there :)
The first 1 is just general
F = ma
The second is the one that "relates the (required) centripetal force to speed and radius."
The 3rd one is angular velocity.
The 4th one is angular velocity squared? I'm not sure what is that.
And the last one...wow. Not even sure what's that either.

But I need a formula I'm sure of :) Thanks to you. I'll use it.

 

[Femme_physics]

Did I get it?

http://img11.imageshack.us/img11/4030/theretherethere2.jpg

Clause B is easy, same as the other exercise getting me back to statics :) That's my thing!

 

[I like Serena]

Did I get it?

Clause B is easy, same as the other exercise getting me back to statics :) That's my thing!

Almost, but there are 2 forces: the tensile force and the force of gravity.
The resultant force must be equal to the centripetal force.

However, this means the tensile force is a little bigger!

 

[I like Serena]

More destruction! More destruction! :D

:D

And we were given this list of formulas.

I can see that formula there :)
The first 1 is just general
F = ma
The second is the one that "relates the (required) centripetal force to speed and radius."
The 3rd one is angular velocity.
The 4th one is angular velocity squared? I'm not sure what is that.
And the last one...wow. Not even sure what's that either.

But I need a formula I'm sure of :) Thanks to you. I'll use it.

Actually, all these formulas identify the centripetal force.

The omegas in between are intermediate steps to find the last formula, which is also the centripetal force, just with different variables.

Note the little r behind a in the first formula F = m a_r
This identifies the centripetal acceleration.


Note in particular that once you introduce T or f, you're assuming it's a uniform circular motion.
Before T and f are introduced the formula is also applicable to non-uniform circular motion (such as a hammer smashing stuff ).

 

[Femme_physics]

Almost, but there are 2 forces: the tensile force and the force of gravity.
The resultant force must be equal to the centripetal force.

However, this means the tensile force is a little bigger!

OH! That's brilliant, I got to say :) Because if the centripetal force was smaller the thing would break. And, if the centripetal force was bigger...well, that's impossible, but the thing would be pulled in closer to the rotation axis.

:) Must... learn... more... (!)

I will probably review this exercise for a long time now (esp. since everybody in class going to ask me about it).

So, ahem...right, the solution! Add 200 [N] you get 1000 [N]! or 999.18

Awesome :)

Thank you! Soooooooo much.

 

[Femme_physics]

Actually, all these formulas identify the centripetal force.

As I'm still not imbued with super level 58 math vision, I'll have to ask: Are you saying all these formulas are essentially the same? Do I really need so many of them then? And by the way, I took a crack at drawing the free body diagrams. Did I get it?

http://img854.imageshack.us/img854/3959/fbd1.jpg

http://img848.imageshack.us/img848/2611/fbd2.jpg

 

[Femme_physics]

Actually, I was just thinking of something when it comes to the free body diagrams:

I think "Vt" (tangential velocity) in drawing is wrong, since at this point tangential velocity turns into a force that hits the sample, so there are two forces at play to add for max tension--> the force of the tangential velocity and the reaction force of the sample. Is that right?

 

[tiny-tim]

Hi Femme_physics!

No, your free body diagram (a) is wrong …

i] the velocity V_t should not be on it at all, it's neither a force nor an acceleration

ii] you must decide which body the diagram refers to: since you're using the acceleration of the hammer (M, at the end of the shaft), the body must be the hammer, so the reaction force N at O is irrelevant: instead you need the tension T, which does act directly on the hammer

iii] you have shown the acceleration the same way as a force: you should show it differently (I refer a double arrow, off to the side of the diagram)

(alternatively, you could use the non-inertial rotating frame of the hammer, in which case the acceleration is zero, and you must instead show the extra, centrifugal force )

… at this point tangential velocity turns into a force that hits the sample, so there are two forces at play to add for max tension--> the force of the tangential velocity and the reaction force of the sample. Is that right?

The reaction force between the hammer and the sample will be horizontal, so it will not affect the tension (which at that point is vertical).

 

[I like Serena]

As I'm still not imbued with super level 58 math vision, I'll have to ask: Are you saying all these formulas are essentially the same? Do I really need so many of them then?

I think that what you need is to understand in how far they are the same and what use they are.

The only thing you need to know of the centripetal force is that:

F_{centripetal} = \frac {m v^2} r

But you also need to know that:

v = \omega \times r

And if the circular motion is uniform:
\omega = \frac {2 \pi} T
f = \frac 1 T

The other formulas can be derived from these.
It may be useful to keep them at hand for easier calculation, but it would help if you see that they are the same.

For instance, if you substitute v from the second formula in the first one, you'll find the other form of the centripetal force: F = m ω² r

 

[Femme_physics]

ii] you must decide which body the diagram refers to: since you're using the acceleration of the hammer (M, at the end of the shaft), the body must be the hammer, so the reaction force N at O is irrelevant: instead you need the tension T, which does act directly on the hammer

Ah! Settings me straight as always :) Of course! If we're looking at the "hammer head" only it emerges "from" the hammer head, not from the support. But wait, the hammer head hasn't been defined to me in terms of length and width. Does it matter?

iii] you have shown the acceleration the same way as a force: you should show it differently (I refer a double arrow, off to the side of the diagram)

Acceleration is always towards the axis of rotation. Of course N would have to point the same way because it's opposite of W. It's normal force... hmm..what am I missing here?

The reaction force between the hammer and the sample will be horizontal, so it will not affect the tension (which at that point is vertical).

Interesting! A great fact. And duly noted :)

But you also need to know that:v = \omega \times r

That I know :)

And if the circular motion is uniform:
\omega = \frac {2 \pi} T
f = \frac 1 T

Also know :)

The only thing you need to know of the centripetal force is that:

F_{centripetal} = \frac {m v^2} r

Aha! Then this is the one that lost me. A very important one too! Thanks, formula master :)

(alternatively, you could use the non-inertial rotating frame of the hammer, in which case the acceleration is zero, and you must instead show the extra, centrifugal force )

You wrote it in small text- any reason? I'll look into it later tomorrow. Heading to sleep now-- thanks for all your great awesome help, tiny-t, ILS! :) You rrrrrrrrrrrrrrrrrock.

 

[I like Serena]

Ah! Settings me straight as always :) Of course! If we're looking at the "hammer head" only it emerges "from" the hammer head, not from the support. But wait, the hammer head hasn't been defined to me in terms of length and width. Does it matter?

No, you don't need length and width of the hammerhead.
We'll treat it as a "point mass"!

Acceleration is always towards the axis of rotation. Of course N would have to point the same way because it's opposite of W. It's normal force... hmm..what am I missing here?

You're right.
I think tiny-tim only means you should make a distinction between force vectors, acceleration vectors, and speed vectors.
A different color would do as well. ;)
Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

You wrote it in small text- any reason? I'll look into it later tomorrow. Heading to sleep now-- thanks for all your great awesome help, tiny-t, ILS! :) You rrrrrrrrrrrrrrrrrock.

This is about the centrifugal pseudo-force, which operates in a so called non-inertial frame, meaning there are special rules to it...

 

[Femme_physics]

Had time to sneak in one more reply :)

We'll treat it as a "point mass"!

Good! They should've mentioned it in the question, come to think of it! Why should it just occur to me naturally that we're treating it as point mass?

You're right.
I think tiny-tim only means you should make a distinction between force vectors, acceleration vectors, and speed vectors.
A different color would do as well. ;)

Sold. I'm convinced. :)

(plus, I like colors!)

Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

There's no need to draw acceleration? Really?

This is about the centrifugal pseudo-force, which operates in a so called non-inertial frame, meaning there are special rules to it...

Oh, this is the psedu-force you were talking about? Hmm...special rules... well, this may be a heavy topic and I'll need a fresh mind. Talk to you two tomorrow! :)

 

[I like Serena]

Had time to sneak in one more reply :)

:D

Good! They should've mentioned it in the question, come to think of it! Why should it just occur to me naturally that we're treating it as point mass?

No reason. :)
It's just that its size does not matter.
You can give it a size if you want, but the answers will come out the same.

Do note that the radius R is, and would be, given from the axis to the center of mass of the hammerhead.

Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

There's no need to draw acceleration? Really?

Here's from wikipedia about FBD's:

"A free body diagram, also called a force diagram, is a pictorial representation often used by physicists and engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types acting on this body. "

That is, in a pure form it should contain only force vectors. ;)
Oh, and note that if you add all the force vectors together, you should get a resultant force that points in the same direction as the acceleration.
Their relationship is:
\vec F_{resultant} = m \cdot \vec a

 

[Femme_physics]

Okay, a fresh new day (and once again fresh new coffee spilled on my papers), I'm ready to see if I truly have this problem figured out, starting with my new free body diagram based on tiny-tim's feedback.

This diagram refers to the forces acting on the hammer.

http://img715.imageshack.us/img715/6567/newfbd.jpg

No reason. :)
It's just that its size does not matter.

Who says "size doesn't matter"? ;)

Do note that the radius R is, and would be, given from the axis to the center of mass of the hammerhead.

I did note that :)

Here's from wikipedia about FBD's:

"A free body diagram, also called a force diagram, is a pictorial representation often used by physicists and engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types acting on this body. "

Ah...thanks for this jewel. That's proper engineering right there.

 

[I like Serena]

Okay, a fresh new day (and once again fresh new coffee spilled on my papers), I'm ready to see if I truly have this problem figured out, starting with my new free body diagram based on tiny-tim's feedback.

This diagram refers to the forces acting on the hammer.

Looking nice! :)
(Again, where's the coffee stain?)

That is, you've drawn N smaller than W.
That doesn't leave much for any net centripetal force does it?
Perhaps the hammerhead will be flying away...
Hope it doesn't impact something else!

And you're right, the speed turns into a force.
Right there you have an entirely new beast called "impulse", which is force times time.
I don't think your studying material covered that yet! :)
Note that we have no information about how great this force would be, although we can say a little on how large the change in "momentum" will be (another, or rather the same, new beast) .

Who says "size doesn't matter"? ;)

Eep! It does?

Ah...thanks for this jewel. That's proper engineering right there.

:)

 

[Femme_physics]

Looking nice! :)
(Again, where's the coffee stain?)

It's my electronics exercise notebook that got the biggest hit. :)

http://img707.imageshack.us/img707/7227/coffeey.jpg

That is, you've drawn N smaller than W.
That doesn't leave much for any net centripetal force does it?
Perhaps the hammerhead will be flying away...
Hope it doesn't impact something else!

Waaaaaaaait a second now, the length of the vector was completely arbitrary. If I knew it mattered, I'd wait for the results before drawing them. But other than the lengths, it's correct right?

And you're right, the speed turns into a force.
Right there you have an entirely new beast called "impulse", which is force times time.
I don't think your studying material covered that yet! :)
Note that we have no information about how great this force would be, although we can say a little on how large the change in "momentum" will be (another, or rather the same, new beast) .

That's a fascinating input!

I love how you expand the engineering realm to me when I study something basic, and you show more something more advanced that I'm able to understand :)

Eep! It does?

Nah, was just looking for a joke ;)

 

[Femme_physics]

Earlier I wrote this:

I used the diameter for "h" -- is that right? it's the change in position, delta x. So, the change in position is 1.6m!

Oh, and am I right in saying that the change in position is 1.6 and not pi? because we're looking at the displacement, if I understand correctly.

 

[I like Serena]

It's my electronics exercise notebook that got the biggest hit. :)

There! That's proper engineering! Showing the results of your actions! ;)

Waaaaaaaait a second now, the length of the vector was completely arbitrary. If I knew it mattered, I'd wait for the results before drawing them. But other than the lengths, it's correct right?

Didn't you think that, before you made your calculations, it mattered whether the hammer would be flying away or not?
All in a fraction of a second that apparently you're waiting for! :)

EDIT: And yes, it is correct, regardless of the lengths.

That's a fascinating input!

I love how you expand the engineering realm to me when I study something basic, and you show more something more advanced that I'm able to understand :)

Any time! :)

Nah, was just looking for a joke ;)

Good!
Because a point mass is a pointy thingy with very little size.
Or rather no size at all!

Earlier I wrote this:

Oh, and am I right in saying that the change in position is 1.6 and not pi? because we're looking at the displacement, if I understand correctly.

Yep!
As tiny-tim already stated, it does not matter by which route you got from A to B, only the distance (in height!) matters in conservation of energy.

 

[Femme_physics]

Didn't you think that, before you made your calculations, it mattered whether the hammer would be flying away or not?
All in a fraction of a second that apparently you're waiting for! :)

Yes, you're right, I should've intuitively realized that. BUT, do we draw vectors longer based on their magnitude, or is it just being pedantic?

Because a point mass is a pointy thingy with very little size.

But it's not really a point mass because its mass is defined, right? So it's a 200[N] point mass, if anything :)

As tiny-tim already stated, it does not matter by which route you got from A to B, only the distance (in height!) matters in conservation of energy.

Brilliant :)

 

[I like Serena]

Yes, you're right, I should've intuitively realized that. BUT, do we draw vectors longer based on their magnitude, or is it just being pedantic?

Yes, we draw vectors longer based on their magnitude.

That brings us to "vector addition".
There are 2 ways to add vectors: geometric and algebraic.

Geometric means that you draw all vectors carefully on scale, and when you draw each vector head-to-tail after each other, you can measure the length of the resulting vector on paper.

Algebraic means that you typically calculate the x component and the y component of each force and that you add those numerically.
I think you're already quite used to that in your mechanics problems.

There's no need to do things geometrically in your case, although it would be nice if what you draw is more or less geometrically correct. :)

But it's not really a point mass because its mass is defined, right? So it's a 200[N] point mass, if anything :)

Hmm, 200 N is not a mass - it's a force, or rather, a weight.

And for a point mass, its mass is always defined. It's just that its size is not defined!

 

[Femme_physics]

Yes, we draw vectors longer based on their magnitude.

That brings us to "vector addition".
There are 2 ways to add vectors: geometric and algebraic.

Geometric means that you draw all vectors carefully on scale, and when you draw each vector head-to-tail after each other, you can measure the length of the resulting vector on paper.

We never had to measure vectors before, or draw them carefully to scale! Sounds like a system I rather not learn for now, but interesting nevertheless :)

Algebraic means that you typically calculate the x component and the y component of each force and that you add those numerically.
I think you're already quite used to that in your mechanics problems.

Yep!

There's no need to do things geometrically in your case, although it would be nice if what you draw is more or less geometrically correct. :)

Duly noted :)

Hmm, 200 N is not a mass - it's a force, or rather, a weight.

And for a point mass, its mass is always defined. It's just that its size is not defined!

Fair enough!

 

[Femme_physics]

Couple of questions.

1) This does not appear to be uniform circular motion anywhere, but rather NON-uniform circular motion. Is that right?

2) I'm a bit confused as to why I go about ADDING 200 [Newtons] IF the normal force is up, and W is down.

3) My F turned out being 799.18 [N]. I presume F is N, right?

Not sure why I should really add 200 to F and not decrease. Does F means "all the vertical forces" or just N?

Posted my latest full calculation of F.

http://img19.imageshack.us/img19/7048/newcalculations.jpg

 

[I like Serena]

Couple of questions.

1) This does not appear to be uniform circular motion anywhere, but rather NON-uniform circular motion. Is that right?

Right! :)

2) I'm a bit confused as to why I go about ADDING 200 [Newtons] IF the normal force is up, and W is down.

3) My F turned out being 799.18 [N]. I presume F is N, right?

No.
F is your net resultant force that must be equal to the centripetal force.

W and N are the actual forces acting on the hammerhead.
They point in opposite directions and their sum must equal the centripetal force.

So: F = N - W = centripetal force

Which means: N = F + W

Not sure why I should really add 200 to F and not decrease. Does F means "all the vertical forces" or just N?

So yes, F means all the forces (vertical or otherwise) and not just N.

 

[Femme_physics]

So yes, F means all the forces (vertical or otherwise) and not just N.

I wish I've been told that before! :) Thanks.

No.
F is your net resultant force that must be equal to the centripetal force.

So "F" IS in fact the centripetal force, not "EQUAL" to the centripetal force, yes?

W and N are the actual forces acting on the hammerhead.
They point in opposite directions and their sum must equal the centripetal force.

So: F = N - W = centripetal force

Which means: N = F + W

Duly noted :) Excellent. I now feel I can solve other non-uniform circular motion problems.

 

[I like Serena]

So "F" IS in fact the centripetal force, not "EQUAL" to the centripetal force, yes?

Isn't that the same thing?

Duly noted :) Excellent. I now feel I can solve other non-uniform circular motion problems.

You know, I believe so too!

 

[Femme_physics]

Isn't that the same thing?

Heh, I guess. I'm being pedantic :)

You know, I believe so too!

!

Thank you very much!

Btw, where does this formula comes from?

So: F = N - W = centripetal force

Is it simply sum of all forces = ma? I fail to see how that formula (sum of all forces = ma) relates force, to weight, to normal force; since it really relates mass to acceleration.

 

[I like Serena]

Is it simply sum of all forces = ma?

Yes. That is always the case.
This is called Newton's second law.

I fail to see how that formula (sum of all forces = ma) relates force, to weight, to normal force; since it really relates mass to acceleration.

First things first, you call it "normal" force - it isn't - it's tensile force (T would have been a better symbol).
Normal force is the force exerted by a surface.

And these are actually 2 formulas.

You have F = sum of all forces = N - W

And you have F = sum of all forces = m a

In statics you usually want the acceleration a to be zero, otherwise your sexy problem might bolt away from you!

 

[Femme_physics]

And these are actually 2 formulas.

You have F = sum of all forces = N - W

And you have F = sum of all forces = m a

Hmm... 2 formulas, that is in fact the same formula?

How does it make sense? And they're meant for different variables? This is a bit confusing, don't you think?

 

[I like Serena]

Hmm... 2 formulas, that is in fact the same formula?

How does it make sense? And they're meant for different variables? This is a bit confusing, don't you think?

Hmm, the first formula is just "vector addition".
The second formula defines the relationship between resultant force and acceleration.

Since they are both equal to the resultant force, they are equal to each other.

 

[tiny-tim]

Hi Femme_physics!

(just got up :zzz: …)

do we draw vectors longer based on their magnitude, or is it just being pedantic?

On a free body diagram I draw them all the same length (if there's room) …

often, you don't know what length they should be anyway.

(of course in a vector triangle the lengths do matter!)

btw, I agree with I like Serena … there's no normal force here, you should use T for tension

So "F" IS in fact the centripetal force, not "EQUAL" to the centripetal force, yes?

I wish people wouldn't say "centripetal force".

In this case, T is a centripetal force and since there's only one, it's the centripetal force … T - W is the net centripetal force.

The correct equation to use is (as almost always) good ol' Newton's second law …

net force = mass time acceleration …​

which in this case is

vertical component of net force = mass time centripetal acceleration ​

(finally, my small-print earlier was because I assumed you hadn't done centrifugal force )