So I'm working on something and I've got myself stumped on part of it that I know is something easy. I know it is basically a really easy dynamics problem, but I keep confusing myself for some reason. I need to relate the lateral motion of the top top point to the downward motion of the bottom point.

can someone please give me some fresh eyes on this

2/27/2010 9:45:49 PM

no

2/27/2010 9:46:06 PM

did you mean to post this in study hall?

2/27/2010 9:47:47 PM

not really, i'm trying to get it done today and no one goes into study hall on saturday night

2/27/2010 9:48:42 PM

a^2+b^2=c^2

2/27/2010 9:49:36 PM

this really isn't hard

2/27/2010 9:52:56 PM

butt sex... butt sex is always the answer to dynamics problems... especially if Chow is your Professor

2/27/2010 9:53:23 PM

also, omega is roughly constant.

so the position of the top i guess is the vector a, and the bottom is the vector c.

i know its not hard, but the last time i did anything having to do with dynamics or vector analysis was 7 damn years ago

[Edited on February 27, 2010 at 9:56 PM. Reason : .]

2/27/2010 9:56:01 PM

maybe…

(a-c)(sin(t-B))=c(cos(t))

t=theta

based on the parametric equations for a circle: http://mathworld.wolfram.com/ParametricEquations.html




[Edited on February 27, 2010 at 10:09 PM. Reason : this is wrong actually...]

2/27/2010 9:56:24 PM

i mean i can relate any of the sides together with the law of sines or cosines, i even labeled the angles and sides according to the conventional definitions.

2/27/2010 9:57:33 PM

speed of each point is proportional to distance from the rotation point. then you just have to deal with the cos/sin to figure out what proportion of that motion is up/down and left/right

2/27/2010 9:58:21 PM

thats the part i'm tripping up on

2/27/2010 9:59:01 PM

which part?

2/27/2010 9:59:45 PM

i get that v=omega*r how do i get x and y scalars?

2/27/2010 10:00:42 PM

use trigonometry to figure out the angle from the horizontal for the two points.

i'm going to do it for fun and then tell you if you get it right.

[Edited on February 27, 2010 at 10:02 PM. Reason : .]

2/27/2010 10:01:29 PM

yeah, thats the part i'm tripping myself up on

2/27/2010 10:02:14 PM

well, you know that the velocity is perpendicular to the radius

[Edited on February 27, 2010 at 10:08 PM. Reason : (theta and theta + beta are the only angles you should care about)]

2/27/2010 10:03:43 PM

don't i need rho to find components of the top part?

2/27/2010 10:10:09 PM

no.

eg at theta = 0 velocity of the top vertex is all vertical. at theta = 90 it's all to the left. the same sort of logic can be used for the other vertex with the angle theta + beta.

[Edited on February 27, 2010 at 10:12 PM. Reason : i got an a+ in dynamics 8 years ago]

2/27/2010 10:10:31 PM

maybe covert it to polar equations, then solve, then break out the x/y components?

2/27/2010 10:10:57 PM

absolutely no need.

2/27/2010 10:12:34 PM

ok, i don't understand where you are going then?

really vector analysis would be the best way with where i have to go next with this, but this part of it was so easy i wasn't going to do that and then i confused myself

2/27/2010 10:12:57 PM

i can't really speak for what you need to do next. but for this part, only theta, beta, a and c are needed (and possibly omega)

i suppose what i'm saying would be polar though. so i guess moron is right. just no real 'converting' to polar is necessary

but there is converting from polar to cartesian for those x/y components

[Edited on February 27, 2010 at 10:15 PM. Reason : .]

2/27/2010 10:14:26 PM

Quote :

"eg at theta = 0 velocity of the top vertex is all vertical. at theta = 90 it's all to the left. the same sort of logic can be used for the other vertex with the angle theta + beta."

thats only true if beta = 90, beta does not necessarily equal 90

(in fact it almost certainly is greater than 90 in my case)

edit":
misunderstood what you were saying, read it as "at the other vertex" its all to the left


[Edited on February 27, 2010 at 10:22 PM. Reason : .]

2/27/2010 10:16:52 PM

no. you are wrong.

2/27/2010 10:17:59 PM

when theta=0 the bottom vertex will be moving down AND to the left

tangent to c

[Edited on February 27, 2010 at 10:19 PM. Reason : .]

2/27/2010 10:18:58 PM

that's why i said that you use theta + beta for the bottom vertex.

^and it would not necessarily be tangent to c

[Edited on February 27, 2010 at 10:20 PM. Reason : .]

2/27/2010 10:19:30 PM

^see my edit, i misread your post

^well i don't understand where you are going with that

[Edited on February 27, 2010 at 10:21 PM. Reason : .]

2/27/2010 10:20:12 PM

forget entirely about the triangle. and treat is as two separate lines. each with an angle from the horizontal

for the top vertex, the angle is theta. for the bottom, let's called that angle lambda. if lambda is 90, the motion is all to the left. if lambda is 180, all down, etc.

now replace lambda with theta + beta.

2/27/2010 10:22:12 PM

A= (1/2)B*H

2/27/2010 10:22:31 PM

^^i understand that, where are you going with that

[Edited on February 27, 2010 at 10:23 PM. Reason : .]

2/27/2010 10:23:01 PM

figure out the horizontal and vertical components of those velocities using conversion from polar to cartesian. and you can relate that because they both use theta in their formulations.

2/27/2010 10:24:22 PM

thats what i cant do, thats where i was when i made my first post

[Edited on February 27, 2010 at 10:25 PM. Reason : ,]

2/27/2010 10:24:52 PM

dear god buddy. your're telling me you can't convert a simple angular velocity to cartesian and you're halfway through a dynamics class?

2/27/2010 10:25:52 PM

i took dynamics in 2003

2/27/2010 10:27:19 PM

http://delphiforfun.org/programs/math_topics/polar-cartesian.htm

[Edited on February 27, 2010 at 10:28 PM. Reason : ^ i took it in 2002 and still remember this crap]

2/27/2010 10:27:44 PM

dude I told you next time do it in the butt

2/27/2010 10:27:54 PM

http://en.wikipedia.org/wiki/Polar_coordinate_system#Converting_between_polar_and_Cartesian_coordinates

2/27/2010 10:29:52 PM

i haven't touched this stuff since then though, i wasnt in an engineering related field

so the angle between the tangent at the top vertex and the x component equals theta?

^i know that

[Edited on February 27, 2010 at 10:32 PM. Reason : .]

2/27/2010 10:32:14 PM

wait. all you need is the x component and y component, no?

Vx = -r*sin(theta)
Vy = r*cos(theta)

[Edited on February 27, 2010 at 10:35 PM. Reason : .]

[Edited on February 27, 2010 at 10:35 PM. Reason : .]

[Edited on February 27, 2010 at 10:38 PM. Reason : .]

2/27/2010 10:33:36 PM

i guess this was what i needed

2/27/2010 10:37:22 PM

no. not at all. you are making this more complicated that it needs to be.

(note my edit, that pic is for position, for velocity a derivative needs to be taken)

[Edited on February 27, 2010 at 10:39 PM. Reason : .]

2/27/2010 10:37:58 PM

nevermind, i'm an idiot

2/27/2010 10:39:09 PM

ax=a*cos(t) (lateral motion of top point)

cy=c*sin(t+b) (downward motion of bottom point)

or

in polar...

motion of a with respect to theta(t):
r(t)=a

motion of c:
r(t+b)=c

2/27/2010 10:41:25 PM

^that's not right. you need a derivative for velocity, Vx = -rsint, Vy = rcost

2/27/2010 10:43:19 PM

I don't see where he needs velocity, i did position



[Edited on February 27, 2010 at 10:46 PM. Reason : ]

2/27/2010 10:44:10 PM

gravity? where is gravity anywhere here. and from what i understand, velocity is all he's concerned about.



[Edited on February 27, 2010 at 10:46 PM. Reason : .]

2/27/2010 10:45:55 PM

i don't know, i just interpreted that as position. and i thought the downward arrow there was gravity...

because for a CPE major who has only taken statics (but did great in py 205), doing that with gravity is a bitch of a problem

[Edited on February 27, 2010 at 10:54 PM. Reason : ]

2/27/2010 10:48:08 PM

no. he wants vertical/horizontal component of velocity. which he said in the first post.

the word "dynamics" is in my job description.

heh. though it's not exactly this sort of thing.

[Edited on February 27, 2010 at 10:50 PM. Reason : .]

2/27/2010 10:48:56 PM

set em up

2/27/2010 10:56:49 PM