Dynamics circular and linear motion

[Femme_physics]

I wasn't even sure how to title this exercise since I don't really know how this movement is being officially called in English, I just called it "circular and linear motion" as a translation from Hebrew.

Homework Statement

http://img842.imageshack.us/img842/2149/xycoordinate.jpg

The uniform beam AC moves so its tip A slides along the y axis, upwards, whereas B slides along the x-axis (see drawing). The speed of point A (Va) equals 7 m/s. The measurements are in meters.

Calculate the speed of point C and B

The Attempt at a Solution

The formulas I've seen used appear to give me the wrong solution.

http://img853.imageshack.us/img853/7669/try1v.jpg

http://img62.imageshack.us/img62/5676/try2v.jpg

 

[I like Serena]

Hi Fp!

I wasn't even sure how to title this exercise since I don't really know how this movement is being officially called in English, I just called it "circular and linear motion" as a translation from Hebrew.

Seem like a fine name to me!

Better than "uniform circular motion of a pendulum" for a hammer that falls down and breaks something.


I think we're finally getting to use derivatives in kinematics.
At least that is what I would suggest doing.

The relevant formulas here would be:

v_B = x_B'

v_A = y_A'

Can you perhaps write x_B as a function of y_A?
And then take the derivative?

 

[Femme_physics]

Hi Fp!

Hi ILS! :)

Seem like a fine name to me!

w00t! Female intuition ftw!

...And then take the derivative?

*shock and awe*

Is there a way to solve it without taking the derivative? We haven't solved an exercise by taking the derivative yet! I'd be shocking my classmates if I did, though :D

But I rather take the non-caulculus route first if I can. Does it exist? Can you direct to that?

 

[Femme_physics]

I also noticed I wrote the wrong angle (typo). My Vb should be correct if I fix the angle, I think :) Yep. Got it. Vb is 2.2 m/s

That's at least for Vb. I doubt I can solve the other stuff so I hope my help won't go anywhere. :)

For Vc, I just take a different angle...now I'm not sure which but let me make an attempt. Thanks :)

 

[I like Serena]

*shock and awe*

Is there a way to solve it without taking the derivative? We haven't solved an exercise by taking the derivative yet! I'd be shocking my classmates if I did, though :D

But I rather take the non-caulculus route first if I can. Does it exist? Can you direct to that?

Well, as yet I can't think of a (reasonable) way to explain it without derivatives.

I also noticed I wrote the wrong angle (typo). My Vb should be correct if I fix the angle, I think :) Yep. Got it. Vb is 2.2 m/s

That's at least for Vb. I doubt I can solve the other stuff so I hope my help won't go anywhere. :)

For Vc, I just take a different angle...now I'm not sure which but let me make an attempt. Thanks :)

Yes, you've got the right answer for Vb, although I don't understand how you got to it.
As far as I can tell you juggled with angles and cosines and happened to hit on the right answer.
Somewhere there must be an explanation for it, but as yet I don't see it.

Actually, I can tell you (based on the derivative method) that:

$$v_B = \frac {y_A} {x_B} v_A$$

I can even reason that out in hindsight without derivative, although for point C it becomes another matter again.

 

[lackos]

As soon as i saw this, i was like implicit differentiation. however to go about it another way is something I am going to have to think about.

 

[Femme_physics]

Really? You guys actually "don't know" something? ^^ *even more shock and awe!*

Well, I guess you just learned how to solve it in a more efficient method.

Me? I'm stuck in a non-calculus period.

I'll post the full solution from the solution manual if it helps you understand

http://img825.imageshack.us/img825/5913/solution1.jpg

http://img35.imageshack.us/img35/9295/solution2c.jpg

 

[I like Serena]

Really? You guys actually "don't know" something? ^^ *even more shock and awe!*

Well, I guess you just learned how to solve it in a more efficient method.

Me? I'm stuck in a non-calculus period.

I'll post the full solution from the solution manual if it helps you understand

Wooooow!

Do you understand how you solved it?

At first look I think it's kind of funny they wrote down a formula containing v₁ and v₂ that do not appear anywhere else!

Anyway, this is a little too complex for me to understand while I'm at my job where I do not have much time. I'll get back to you later!

 

[Femme_physics]

Thanks ILS! By the sound of it, maybe taking the derivative IS tons easier! I'll try it tomorrow just to surprise my teacher and classmates. Would be very thrilled if you help me understand it! :)

 

[I like Serena]

I understand a little bit of your solution manual now.

They found the component of v_A that is along the length of the rod.
Same thing for the component of v_B along the length of the rod.

Since the rod wouldn't become any longer or shorter these 2 components must be the same.

And so their v₁ should be v_A, and their v₂ should be v_B.

As for point C, they've lost me again.
I would be able to puzzle it out, but it'll take me some time...

Thanks ILS! By the sound of it, maybe taking the derivative IS tons easier! I'll try it tomorrow just to surprise my teacher and classmates. Would be very thrilled if you help me understand it! :)

Sure! Anything for you!

So let me repeat my question.
Can you find a formula for x_B as a function of y_A?

 

[Femme_physics]

I would be able to puzzle it out, but it'll take me some time...

Well, you don't have to figure it out if it's too much work! But, as they always say, when you teach you're taught ;) Even the very great have something to learn! :D

Sure! Anything for you!

:shy:

Can you find a formula for xB as a function of yA?

Yes, but give me the day off please. ^^ Tomorrow I'll go full speed and will be online for most of the day till class :)

Thank you!

 

[I like Serena]

Yes, but give me the day off please. ^^ Tomorrow I'll go full speed and will be online for most of the day till class :)

The day off?
Why?
It isn't as if you've been up and running for 15 hours yet!

 

[Femme_physics]

Heh ;)

Well it's a new day today, and I'm ready to tackle that fancy derivative method, so not only I'd have another view of understanding, but impress my classmates as well!


BTW- I met with my lecturer yesterday and he told me that he didn't plan us to solve those exercises yet! :D Only today we're supposed to have the lesson about this exercise, but I rather come ready, and have the derivative PoV of the thing.

So am now looking at this:

$$v_B = \frac {y_A} {x_B} v_A$$

And am thinking to myself to write the functions first that you asked me in your original reply.

So let me repeat my question.
Can you find a formula for x_B as a function of y_A?

[/quote]

I can try!

http://img802.imageshack.us/img802/3020/thefunctions.jpg

 

[I like Serena]

Heh ;)

Well it's a new day today, and I'm ready to tackle that fancy derivative method, so not only I'd have another view of understanding, but impress my classmates as well!

:)

BTW- I met with my lecturer yesterday and he told me that he didn't plan us to solve those exercises yet! :D Only today we're supposed to have the lesson about this exercise, but I rather come ready, and have the derivative PoV of the thing.

Well, I figured out how they got to point C as well...

So am now looking at this:

$$v_B = \frac {y_A} {x_B} v_A$$

This is the result that we'll get when we are done with the derivative method for point B.

And am thinking to myself to write the functions first that you asked me in your original reply.

I'm afraid I didn't make myself clear enough. :)

I meant you to write 1 equation with only y_A and x_B as variables.
And with y_A I meant the distance of point A to the origin (OA), which is in fact the y coordinate of point A.
With x_B I meant the distance of point B to the origin (OB), which is the x coordinate of point B.

Can you tell how distance between points A and B is related?

 

[Femme_physics]

Well, I figured out how they got to point C as well...

It all comes down to the instantaneous axis of rotation, right?

This is the result that we'll get when we are done with the derivative method for point B.

Alright, then step by step, beat by beat :)

With xB I meant the distance of point B to the origin (OB), which is the x coordinate of point B.

Where's point O, though? Here?

http://img64.imageshack.us/img64/4339/pointo.jpg

Can you tell how distance between points A and B is related?

By the hypotenuse, right?

 

[I like Serena]

It all comes down to the instantaneous axis of rotation, right?

Yep!

Alright, then step by step, beat by beat :)

Yep!

Where's point O, though? Here?

Yep!

By the hypotenuse, right?

Yep!

 

[Femme_physics]

I love hearing so many yes's row after row :D

Well, this is how A and B are related, I think:

http://img228.imageshack.us/img228/9655/xbya.jpg

Is that it?EDIT: hmm, wait, come to think of it, the hypotenuse should be a variable too, no?

 

[I like Serena]

Well, this is how A and B are related, I think:

Is that it?

EDIT: hmm, wait, come to think of it, the hypotenuse should be a variable too, no?

This would be the square of the hypotenuse.
But you introduced alpha in the expression, but we don't want alpha in it...

And actually, the hypotenuse is given, being 5.
And Pythagoras applies here...

 

[Femme_physics]

Hmm, maybe I'm jumping ahead of myself trying to do the derivative thing. If my lecturer says it's not necessary then I think I'll leave it be and won't try to brag new solving methods that seems to be ahead of me! The solution manual's method is a lot more simple to me :) and through some "reverse engineering" I am able to understand it. I'll stick with my classmates and give them my lecturer's method :)

Thanks for the effort invested in me, ILS! I'm just too much of a calculus knucklehead to figure it out.

I'd just like to ask one question if I may. Why is this P not the axis of rotation, and the other one is? Is it because where our movements are defined?

http://img809.imageshack.us/img809/2751/axisofrotation.jpg

 

[I like Serena]

Thanks for the effort invested in me, ILS! I'm just too much of a calculus knucklehead to figure it out.

Ah, I was trying to get you to write down:

$$x_B = \sqrt {5^2 - {y_A}^2}$$

But I guess I shouldn't have been so difficult about it. Sorry! :shy:

I'd just like to ask one question if I may. Why is this P not the axis of rotation, and the other one is? Is it because where our movements are defined?

P is supposed to be the center of a circle around which A, B and C are revolving.
In particular the speed at any point of a circle is tangential to the circle (do you know what I mean with "tangential"?).

For the speed at B to be tangential to the circle it's on, point P must be straight above B.

Similary point P must be straight right from A.

 

[Femme_physics]

But I guess I shouldn't have been so difficult about it. Sorry!

No no, you're sticking with your teaching style no matter what, and it's great! :) I just have too much of a hole solving physics with calculus and I guess you didn't expect that! How could you? I'll stick to what I do best for now-- non-calc physics!

P is supposed to be the center of a circle around which A, B and C are revolving.

Yes, I understand that.

In particular the speed at any point of a circle is tangential to the circle (do you know what I mean with "tangential"?).

Yes, like tangential velocity, it's the vector that causes the rotational movement. It has to be 90 degrees from the axis. If it's not, some energy is being wasted pressing or pulling the axis.

For the speed at B to be tangential to the circle it's on, point P must be straight above B.

Similary point P must be straight left from A.

Fair enough. Then how come when calculating point C we're not using the axis of the pic I posted? It's "must be straight above C", no?

 

[I like Serena]

No no, you're sticking with your teaching style no matter what, and it's great! :) I just have too much of a hole solving physics with calculus and I guess you didn't expect that! How could you? I'll stick to what I do best for now-- non-calc physics!

Well, it's strange, but sometimes you pick up all the calculus concepts one after the other, like you're on a calculus marathon.
And sometimes it seems as if I'm just not reaching you, as if you're somehow elsewhere with your mind.

I think it has something to do with the abstraction level.
If it looks abstract, it's as if you stop thinking.
But if you can visualize it, and know what you're doing, you've got no problem at all.

I guess the challenge for me is to bring it in a fun way that you can visualize and that "lives" for you.
Yes! I'm learning as well!

Yes, like tangential velocity, it's the vector that causes the rotational movement. It has to be 90 degrees from the axis. If it's not, some energy is being wasted pressing or pulling the axis.

Right!

Fair enough. Then how come when calculating point C we're not using the axis of the pic I posted? It's "must be straight above C", no?

Uhh, no.
Point A is constrained to move over the y axis, so its speed is vertical.
Point B is constrained to move over the x axis, so its speed is horizontal.
Point C is not constrained in such a manner, so its speed will be at an angle, but since you can find point P from A and B, you can deduce the angle of the speed at C.

 

[Femme_physics]

Well, it's strange, but sometimes you pick up all the calculus concepts one after the other, like you're on a calculus marathon.
And sometimes it seems as if I'm just not reaching you, as if you're somehow elsewhere with your mind.

I think it has something to do with the abstraction level.
If it looks abstract, it's as if you stop thinking.
But if you can visualize it, and know what you're doing, you've got no problem at all.

I love it how you're describing my psychology, it's beautiful, and it shows how invested you are in wanting to reach to people. Amazing! You should be a teacher full time not a software engineer!

Yes, using calculus right now seems abstract in physics. As long as I don't need it, I'll drop it.

I guess the challenge for me is to bring it in a fun way that you can visualize and that "lives" for you.
Yes! I'm learning as well!

You're great! I don't know if you should change anything. Perfection shouldn't be replanned ;)

Uhh, no.
Point A is constrained to move over the y axis, so its speed is vertical.
Point B is constrained to move over the x axis, so its speed is horizontal.
Point C is not constrained in such a manner, so its speed will be at an angle, but since you can find point P from A and B, you can deduce the angle of the speed at C.
__________________

Brilliant explanation! Now I understand :)