Uniform circular motion with a pendulum

[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.
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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!