Uniform circular motion with a pendulum

[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 )

 

[Femme_physics]

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.
__________________

Hmm..fair enough. Not sure I 100% get it, but I appreciate the effort :)

Can you please tell me if my notation is correct:

http://img225.imageshack.us/img225/1707/thesteps1.jpg

http://img580.imageshack.us/img580/4421/thesteps2.jpg

Hi Femme_physics!

And W and Fc are in absolute value anyway (I think), since it's the net force

(just got up

Hi tiny-tim!

Good morning ;)

The correct equation to use is (as almost always) good ol' Newton's second law …
net force = mass time acceleration …

By net force do you mean the absolute value of all the forces acting on the axis?

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

Hmm. Why's that? Is there something improper in saying that?

 

[tiny-tim]

By net force do you mean the absolute value of all the forces acting on the axis?

I mean the sum of all those forces (the resultant force, or total force).

Hmm. Why's that? Is there something improper in saying that?

It is confusing you because calling it a force means that you think of it as a separate force, and you want to put it on the LHS of Newton's second law with the other forces … that's double-accounting!

Instead, use mass time centripetal acceleration , and put it on the RHS where it belongs …

that way you won't have problems with the sign.

Centripetal force is not a separate force.

It is only an alternative name for the radially inward component of tension or friction or other force or forces on a body.

See the PF Library on centripetal force for a full discussion.
​

 

[I like Serena]

Can you please tell me if my notation is correct:

http://img225.imageshack.us/img225/1707/thesteps1.jpg

Let's see.
I see a few inconsistencies...
You're showing that you use the formula g h = v² / 2.
(You may want to add a comment that this is about conservation of energy. Not everyone has the super math/physics vision feat! )

But then you write "Δx".
That's not right for 2 reasons.
It implies it's a horizontal distance (by the use of x), but it's a vertical one.
And you just wrote a formula with "h" in it, so you should use "h" instead of "Δx".
Furthermore you write m = 200 [N].
But m is supposed to be a mass and not a force.
So this should be: m = (200 / 9.81) kg.

Then a little further down you refer to the same mass with a capital "M".
You should use the same symbol, that is, "m".
(When you're referring to the weight you'll use W = 200 N.)
In your formula for Fc you have substituted v, but you seem to have forgotten to square it.
Sorry for being such a nitpicker!

 

[Femme_physics]

And you just wrote a formula with "h" in it, so you should use "h" instead of "Δx".

Or delta y?

Furthermore you write m = 200 [N].
But m is supposed to be a mass and not a force.
So this should be: m = (200 / 9.81) kg.

Well, that's just a given value, I alter it in my equations to kg or Newtons as needs be :) makes sense no?

Then a little further down you refer to the same mass with a capital "M".
You should use the same symbol, that is, "m".
(When you're referring to the weight you'll use W = 200 N.)

Noted and checked.

I like your nitpicking, keep it up :)

In your formula for Fc you have substituted v, but you seem to have forgotten to square it.

True, indiscretion, but I did write the squared result for that.

Sorry for being such a nitpicker!

I love nitpickers! :)

 

[Femme_physics]

And while I'm at it: is mass times acceleration always a positive value? How did I know not to write it in a minus in this case?

 

[I like Serena]

http://img580.imageshack.us/img580/4421/thesteps2.jpg

I love nitpickers! :)

All right! Since you apparently like it, I'll continue with your next picture.

You write here the formula that the sum of the forces is equal to mass times acceleration.
(Newton's good ol second law. )
But then you write the forces in the wrong order.

Going along with tiny-tim, it should be:
\sum F = m a
T - W = m \cdot a_{centripetal}

with
a_{centripetal} = \frac {v^2} R

Or delta y?

Yep! That's a good one too!

Well, that's just a given value, I alter it in my equations to kg or Newtons as needs be :) makes sense no?

Uhh, no.
Mass is different from weight, which is a force.
It's not a conversion of units. These are different physical entities.
If you want to show the given value, you should write: W = 200 N.

And while I'm at it: is mass times acceleration always a positive value? How did I know not to write it in a minus in this case?

Hmm, not sure I understand your question.

What you have is that mass is always positive.

Acceleration is a vector that you represent by a value.
That value can be positive or negative, but if it's negative that means that the acceleration vector is actually opposite to the direction you've drawn it.

 

[Femme_physics]

I did fix my sum of all forces = ma shortly after I made that post, and realized my error, thanks :)

Uhh, no.
Mass is different from weight, which is a force.
It's not a conversion of units. These are different physical entities.
If you want to show the given value, you should write: W = 200 N.

I do realize mass is different than weight. Wait, I'm confused. Can we say that while you're obviously always right you can let me off the hook this time on the basis of "not super relevant" as the other stuff I'm not knowing atm?

Hmm, not sure I understand your question.

What you have is that mass is always positive.

Acceleration is a vector that you represent by a value.
That value can be positive or negative, but if it's negative that means that the acceleration vector is actually opposite to the direction you've drawn it.

I see. Let me see how it applies on other problems. Thanks :)

 

[tiny-tim]

And while I'm at it: is mass times acceleration always a positive value? How did I know not to write it in a minus in this case?

I don't understand this question either.

The RHS of F_total = ma is always ma (not -ma).

 

[Femme_physics]

Fair enough! :)

Thanks tiny-tim, ILS! my heroes :)

I'll scan my full solution with ALL the corrections in the evening

 

[I like Serena]

I do realize mass is different than weight. Wait, I'm confused. Can we say that while you're obviously always right you can let me off the hook this time on the basis of "not super relevant" as the other stuff I'm not knowing atm?

Let you off the hook? Let you off the hook?
I'm trying to get you hooked!

But oh well, if you really want to be off the hook, we can let it go as being "not super relevant".
Especially in an engineering environment where most people are not nitpickers. :)
Engineers usually just want the results.

Just do better in the future!