dynamics . joint kinematics

[ahmeeeeeeeeeed]

Hello ,

 

here is a problem

 

http://www.flickr.com/photos/76599498@N07/7070964675/

 

when I solve it , the pin turns to be going upwards , which I can't find logical , what do you see ,please ?

 

here's my answer

 

http://www.flickr.com/photos/76599498@N07/7070966251/

 

http://www.flickr.com/photos/76599498@N07/7070966875/

[ahmeeeeeeeeeed]

???????????

[imatfaal]

Ahmeeed - I think you might get a better response if your posts were more than just photos of your text book and v rough notes. Yours questions are a bit beyond me and I think you are probably putting off some of the heavy-hitters in maths with your presentation

 

Could you look at it this way? Change the question - if the pin was moving within the groove and driving the circular arm how fast would the circular arm move? Surely that would be just the horizontal component of the angular velocity? Can you not reverse that idea

[Xittenn]

I don't understand the wording of the question. Is A stationary? Is B the pin? Is the question asking for [math] \omega [/math]? Are you expected to use polar coordinates--I ask because I am not quite at intermediate mechanics and am expected not to use polar coordinates. Is the groove a blow up of the inside of the block? These may be obvious but if you are seeking help, then maybe not. All things aside the question isn't anything that can't be solved from basic fundamental principles . . . if the block is moving at 60m/s, then I don't see why the x velocity of the pin shouldn't be the same value and the rest should come pretty quickly!

Edited April 16, 2012 by Xittenn

[imatfaal]

Xittenn (youre right, it is a long way from clear) - I am assuming that Point A is fixed, as it the 'housing' of CD. CD is linked to AB by a pin at B (perpendicular to plane of the diagram) which is slotted into a groove running a circular course around the C end of CD. As CD moved to diagram Right the pin at B will be forced to the point where Phi is zero and A B and D all lie on a single line (the dotted one on the diagram). The velocity of the pin at B will always be tangential to the circular groove and will be equal to the angular velocity times the radius of the circle. Its the same problem as the crank shaft that Ahmeeed posted before - but in reverse; if you moved B on a circular course (ie varying Phi) you would either push A to diagram left (similar to crank problem) or if A is fixed then you move CD to the right.

 

The velocity of CD is the rate of change of the distance between A and CD.

 

The Distance between A and CD is in three parts.

- Let O be the point on line AD that the Phi is measured from - ie the centre of the section of circle

- Let M be the point on line AD that will form two rightangled triangles ABM and OMB

The total distance is AM + MO + unknown/unchanging distance to D. The unknown distance can be dropped

 

We know that BM is OB.Sin(Phi) and similarly MO is OB.Cos(Phi).

By Pythagoras we now can say that AM is Sqrt(AB^2-BM^2)= Sqrt(AB^2 - OB^2.Sin^2(phi))

 

The total distance AO is thus AM + MO = OB.Cos(Phi) +Sqrt(AB^2 - OB^2.Sin^2(Phi))

[ahmeeeeeeeeeed]

Ahmeeed - I think you might get a better response if your posts were more than just photos of your text book and v rough notes. Yours questions are a bit beyond me and I think you are probably putting off some of the heavy-hitters in maths with your presentation

 

thanks for response , I put photos of my text book just to let you see the question just as I see it , and for my notes I'll explain it as ( sorry , I was answering just as I would answer in the exam.

 

 

the question is asking for the velocity of the pin (B) at the instant (phi) equals 45 degrees

 

 

the pin is B , and is attached to the arm AB , and allowed to move just in the circular groove , the circular groove itself moves in the (i cap) direction with a constant velocity = 60 m/s

 

as the X velocity is constant = 60 m/s , the circular groove and also the pin have it

 

so what I'm seeking is the velocity of the pin tangential to the circle ...

 

 

I treated this problem as a joint kinematices problem , so I devided it into two parts

 

the first part , I'm treating the pin only as a part of arm( AB) , and so I used polar co-ordinates R cap and theta cap

 

where V = dR/dt in the direction of R cap , and R* omega in the direction of theta cap = (.3)* omega

 

since the length of arm AB is constant then dR/dt = zero and the (theta cap) velocity is R * omega

 

We also have the X velocity of the pin

 

so

 

V = (.3 * omega) theta cap + 60 (i cap) ----------------> equation one

 

 

 

-----

 

secondly I treated the pin as part of the circular part

 

 

we see that both the groove and the pin move in the positive x direction with the same speed , so we can neglect this speed and consider the motion of the pin just circular motion

 

I chose intrinsic co-ordinates with ( tcap) tangential and (n cap) pointing towards the center , I could also use polar again with this .

 

We see that V in the direction tangential in the only velocity compnent here , and it is what I seeking

 

 

then I resuluted the unit vectors (t cap) and (i cap) in the directions of R cap and theta cap using the geomitry of the figure , inorder to equate like quantities , that resulted in the tangential speed equals (- 61.5) t cap

 

which means opposite in sense to the unit vector (t cap) , which also means that the pin is going up ! .. I can't find this logical as i think under this situation the pin should go down the groove not up the groove and here came my question here .

 

 

Yours questions are a bit beyond me and I think you are probably putting off some of the heavy-hitters in maths with your presentation

 

sorry I can't understand this " putting off some of the heavy hitters in maths"

 

 

I don't understand the wording of the question. Is A stationary? Is B the pin? Is the question asking for \omega ? Are you expected to use polar coordinates--I ask because I am not quite at intermediate mechanics and am expected not to use polar coordinates. Is the groove a blow up of the inside of the block? These may be obvious but if you are seeking help, then maybe not. All things aside the question isn't anything that can't be solved from basic fundamental principles . . . if the block is moving at 60m/s, then I don't see why the x velocity of the pin shouldn't be the same value and the rest should come pretty quickly!

 

Yes , A is stationary , B is the pin , the question as asking for the tangential velcoity of the pin , but of cource getting the \omega of the pin in the half -circular groove would give us the tangential speed ..

 

I'm using polar co-ordinates for arm AB as i said above..

 

yes the x velocity of the pin is the same value (60 m/s) but I'm seeking the tangential velocity also.

 

As CD moved to diagram Right the pin at B will be forced to the point where Phi is zero and A B and D all lie on a single line (the dotted one on the diagram)

 

 

that's what I expect , but my way of answering showed that at the instant phi = 45 the tangential velocity well be up not down , showing that phi ( at that instance) is seeking to increase not decrease , which I think is wrong and that's why I asked the question from the beginning ...

 

 

 

---------------------------

 

 

so what I am seeking now is

 

 

would you please tell me how I could find the tangential velocity of the pin B at the instant phi = 45 degrees from the givings in the textbook ? I want both its magnitude and direction , and I really want to help me find where is the mistake in my calculations ?

 

thanks in advance

 

ts the same problem as the crank shaft that Ahmeeed posted before

 

 

well this is not mine if this is what you mean

 

http://www.scienceforums.net/topic/65674-crank-slider-mechanism-velocity-of-piston-angular-velocity-of-link-hp-needed/

 

Edited April 17, 2012 by ahmeeeeeeeeeed

[ahmeeeeeeeeeed]

?????????

[Xittenn]

I thought you were solving the whole problem from some given point in polar coordinates. I'm just not used to addressing this type of system as polar. I also see that [math] \frac{dx}{dt} [/math] is a constant, so maybe it isn't best to approach it as a harmonic oscillator. It's been ten days, I don't know if this question can still be classified as homework at this point? I would think it has now simply passed into the territory of I can't answer the damn question . . . I don't know how to handle this. :/

 

Here's what I did to solve the problem:

 

1) [math] r = r [/math]

 

2) [math] x = r\cos{\phi} + \frac{dx}{dt}t [/math]

 

3) use Pythagoras to form a relation between [math] \phi [/math] and [math] x [/math]

 

4) find the derivative which will give you the constant [math] \omega [/math]

 

5) Apply the mechanical advantage--gear type

 

6) [math] v_{\perp} = \frac{\omega_{pin}}{r_{pin \; arm}} [/math]

 

It's late if I made a mistake I'm sorry, I've never done a question like this before and I'm half asleep . . . . :/

Edited April 28, 2012 by Xittenn