How am I supposed to know the direction of this velocity?

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The discussion revolves around understanding the direction of velocity in a mechanical system involving a rotating shaft and hydraulic cylinder. The user is preparing for a pre-finals test and seeks clarification on the velocity direction of point A (Va) and point B (Vb) in relation to their angular velocities. Key points include the necessity of recognizing that point A moves in a circular motion around a fixed point, which dictates that its velocity is always perpendicular to the radius. There is also a debate about the correctness of formulas used for calculating angular velocity, with users sharing insights on how to properly account for directionality in their calculations. Ultimately, the conversation highlights the importance of understanding the relationship between angular velocity and linear velocity in mechanical systems.
Femme_physics
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(((Since my pre-finals test is Friday, this Friday, I decided to skip the energy questions and go over the basic dynamic energy-less stuff since I'll never get enough practice on energy on time. Please excuse me. I'll return to solve my energy-spring topic before that at a later occasion.

)))

http://img202.imageshack.us/img202/2360/hhheyn.jpg I'll quote the question:

"Controlling the rotation of shaft OA around point O, in the described structure, is performed by analyizing the horizontal motion of hydraulic cylinder shaft piston H, in the described position. The angle, theta, equals 66 degrees. The speed of point B (Vb) is 3 meters per second, and OA is at horizontal position.

---

Now,
The manual shows Va pointing vertically down (no angle to the Y axis). My question is how am I supposed to know this is the case? That this is really the direction of Va's velocity?
 
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Femme_physics said:
Now,
The manual shows Va pointing vertically down (no angle to the Y axis). My question is how am I supposed to know this is the case? That this is really the direction of Va's velocity?

Hi Fp! :smile:


Point O is a fixed point, and OA can not bend.
This means that point A can only make a circular motion around point O.
In a circular motion the speed is always perpendicular...

(Btw, what you do and when or if you do anything is always entirely up to you. ;))
 
Point O is a fixed point, and OA can not bend.
This means that point A can only make a circular motion around point O.
In a circular motion the speed is always perpendicular...

That about clears everything. Thanks :)
 
Femme_physics said:
That about clears everything. Thanks :)

You're welcome and success with your pre-finals test! :smile:
 
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Femme_physics said:
Thanks^^

I'm getting the wrong figure for omega AB. Can you see why?

How did you get to the formula |v_A \sin 30 - v_B \sin 66| = \omega_{AB} \cdot L_{AB}?
It doesn't look right...
 
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Purrrrrrrr! :)

Still, how did you get the formula?
 
  • #10
Femme_physics said:
Oh from my formula sheet

Oh ok. That does seem right. :smile:

Then could you plug the numbers in and calculate omega again?

Edit: For reference, what is your solution manual's value?
I calculated it in a different manner and got omega = 16.7 rad/s

Edit2: Hold on! Your formula sheet is off!
As it is drawn, the formula should be: |v_A \sin 24 + v_B \sin 66| = \omega_{AB} \cdot L_{AB}
 
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  • #12
Femme_physics said:
Same result :(

Edit: Solution manual's value is 16.68 rad/sec. Typical, you'r right :)

Just edited my previous post a couple of times.
Now I think the solution manual's is wrong... :wink:

It should be omega = 18.2 rad/s
 
  • #13
Wow, really? The formulas are wrong? Can you please verify it for sure? I'll send it to the entire class. It's one day before the test! OMG, this is really important!

Does anyone know the English name of this formula??!?
 
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  • #14
Femme_physics said:
Wow, really? The formulas are wrong? Can you please verify it for sure? I'll send it to the entire class

Yes I'm sure.

I can even explain.

Let's start with a stationary stick - it's angular velocity omega is zero.

Now let one end turn with vA - it will have an angular velocity now.

Now let the other end turn in the opposite direction with vB - the stick will turn even faster!
Meaning a higher angular velocity!
 
  • #15
That can't be right. I'm getting the same answer as the manual now using the formula with plus. I see also what they did, they just used Vb in minus sigh so the minus and minus canceled themselves for a plus

http://img232.imageshack.us/img232/4309/solman.jpg
 
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  • #16
Femme_physics said:
Wow, really? The formulas are wrong? Can you please verify it for sure? I'll send it to the entire class. It's one day before the test! OMG, this is really important!

Does anyone know the English name of this formula??!?

I don't think the formula has a name, but I'd call it something like: angular velocity of a moving stick. :)

Note that the formula is not wrong in itself - it's the choice for beta.
If beta (and vB) had been drawn on the other side of the stick, the formula would be correct.
 
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  • #17
Femme_physics said:
That can't be right. I'm getting the same answer as the manual now using the formula with plus. I see also what they did, they just used Vb in minus sigh so the minus and minus canceled themselves for a plus

Yes. You're right. omega = 16.67 rad/s
I mixed up the 180 mm with the length of the stick.
I did it right the first time, but I got excited! ;)

My point remains that the drawing and the formula on your formula sheet don't match.
 
  • #18
Btw, if you're interested, I can give you my method to calculate omega.

It fits in 2 short sentences and 1 simple formula (without trig). :smile:
 
  • #19
I like Serena said:
Yes. You're right. omega = 16.67 rad/s
I mixed up the 180 mm with the length of the stick.
I did it right the first time, but I got excited! ;)

My point remains that the drawing and the formula on your formula sheet don't match.

Ah, good :)

I still don't see how the drawing doesn't match the formula. they just used (-sin(beta)) because the vector points in the negative direction. (our x-axis is the beam). So it makes perfect sense no?



I like Serena said:
Btw, if you're interested, I can give you my method to calculate omega.

It fits in 2 short sentences and 1 simple formula (without trig). :smile:

I'm good with trig! I'm good at trig! Is your method calculusy? Problem is that I want to stick with the entire class and don't want to do the fancy stuff you university grads do! Although I'm curious now, what is it?

I do have another question for a different exercise that I actually want to post in this topic because they relate. Or do you want to give me your method first?
 
  • #20
Femme_physics said:
Ah, good :)

I still don't see how the drawing doesn't match the formula. they just used (-sin(beta)) because the vector points in the negative direction. (our x-axis is the beam). So it makes perfect sense no?

What you say makes perfect sense yes! :smile:

It's just not what's in the drawing.

How are you supposed to know that you have to add a minus sign?
It's not something you should gamble about, or do by trial and error. ;)
There's a reason you did it wrong at first (not you!).




Femme_physics said:
I'm good with trig! I'm good at trig! Is your method calculusy? Problem is that I want to stick with the entire class and don't want to do the fancy stuff you university grads do! Although I'm curious now, what is it?

I do have another question for a different exercise that I actually want to post in this topic because they relate. Or do you want to give me your method first?

No, it's not calculousy!
It's really simple and it fits right in with what you already know and practiced.

But you'll have to do a bit better than that if you want it! :devil:
 
  • #21
How are you supposed to know that you have to add a minus sign?

Well, since I know the direction of Vb I know the Va must go down so the shaft would rotate and not bend. That's my logic, anyway, no?

But you'll have to do a bit better than that if you want it!

Fair enough, didn't mean to dismiss you on that it's just that my test is tomorrow and helping me so splendidly with my priority questions is best :) (did I mention I really REALLY appreciate it?). Tight on time :(

Here's the new thing. Basically you have shaft AB (see the solution manual drawing below) rotating around O at constant velocity of 12 rad/sec, clockwise. I need to find Va and Vb.

http://img171.imageshack.us/img171/618/omegal.jpg

http://img864.imageshack.us/img864/1069/omega2vog.jpg
The solution manual did 0.3 x 12. Why, why not 0.375?

http://img3.imageshack.us/img3/8868/solman222.jpg
 
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  • #22
Sorry, but I really have to go to work now.
I'm already very late because I enjoy this time with you so much! :blushing:

See you later!
 
  • #23
I'm more :)

Fair enough ;) Thanks for everything! <3
 
  • #24
You can look either at OA turning around O, or you can look at AB turning around P.
You seem the have mixed those up.

In the first case: v_A = \omega_0 \times OA

In the second case: v_A = \omega_{AB} \times PA


Note that what they did here, is exactly what I meant for the previous problem! :devil:
 
  • #25
That's exactly where I got stuck, but I fully understand now :) Thhannnnnnnk youuuuuuuuuuu! I need to remind myself that different members have different angular velocities! *smacks forehead*


Wow, I was able to solve about 5 similar problems now. :) I even realized why they used a minus and a plus in this formula:

http://img854.imageshack.us/img854/1857/herel.jpg

They simply relate the forces to the Y axis :) If the forces on y is negative, you open brackets and put the thing as negative, and if it's positive, also open brackets and put it as positive. I just leave the minus sign in the formula and the results always turn out correct. I actually tried to just use plus on a certain problem and got the wrong answer - that's how I figured out I was wrong. The problem is question is this:

http://img818.imageshack.us/img818/457/taltal.jpg

Can't use plus there
 
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  • #26
Femme_physics said:
That's exactly where I got stuck, but I fully understand now :) Thhannnnnnnk youuuuuuuuuuu! I need to remind myself that different members have different angular velocities! *smacks forehead*

:smile:
Femme_physics said:
Wow, I was able to solve about 5 similar problems now. :) I even realized why they used a minus and a plus in this formula:

They simply relate the forces to the Y axis :) If the forces on y is negative, you open brackets and put the thing as negative, and if it's positive, also open brackets and put it as positive. I just leave the minus sign in the formula and the results always turn out correct. I actually tried to just use plus on a certain problem and got the wrong answer - that's how I figured out I was wrong. The problem is question is this:

Can't use plus there

Yes, this is a case where both speeds and angles are on the same side of the stick, meaning it doesn't quite match the drawing.

Either way, I think you've got it under control! :cool:
 

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