Finding the instantaneous axis of rotation (dynamics - circular and linear motion)

Femme_physics

1. The problem statement, all variables and given/known data



Shaft's AC movement is regulated by runways A and B. The angular velocity of the shaft is 3 rad/sec, counterclockwise. When angle psi = 40 degrees. Calculate:

A) The velocities of pistons A and B (Va, Vb)
B) The velocity of point C (Vc) - magnitude and direction towards the horizontal axis.


3. The attempt at a solution

My problem is finding P - the instantaneous axis of rotation. I picked this:




Picking it, I got the correct Va, the correct Vb, and the correct Vc. (I'm still working on the angle of Vc to the horizontal axis - so far it's not correct but maybe I'm doing something obviously wrong.)

And the solution manual picked that:



Are they both correct?

Quinzio

There is a very straightforward way to relate Vc to Va and Vb.
The book solutions are correct but they're complicated

I like Serena

The instantaneous axis is the center of movement.
This means that all points of the body make a circular motion around this axis.

In your case you know the direction of the speed at 2 points.
Are they making a circular motion around your P?

Femme_physics

Well I just wanna know how come P is in the opposite direction to my P? It's pretty easy to solve this the way the manual did actually. You just find the distance from point B to P and the distance from point A to P and the distance from point C to P and do angular velocity times this distance and you get the answer. Finding the angle of C is a bit more tricky, but I first of all wanna know if I got the correct P because it gives me the same result as they....

Edit: Nevermind, I get it. It depends on the direction of the rotation! :)

I like Serena

Hmm, I would have worded it differently.

I'd have said that any speed vector must be perpendicular....

Femme_physics

Fair enough, but if my P is where the solution manual says, then I get opposite results for Va and Vb to what the manual says I should get!

I like Serena

Perhaps if you switched the letters A and B around?

You might say of course that you're not allowed to do that.
But then, if I look at the original problem, it seems to me that they did exactly that!

Femme_physics

*an embarrassing look*

Oh.

*slaps forehead*

Ouch.
I really gotta stop doing that...

Thanks ILS :) Problema el solva (I have no idea if what I wrote is lingually correct)......

I like Serena

What? The switching around of letters and similar stuff?
Or the slapping?

You can slap me if you want!


If it is or not, it sounds nice!

Femme_physics

Actually, still one problem. Finding the angle of Vc to the horizontal axis. I get 61.4 but the manual gets 59.21... I feel it's too much to chalk it up to "rounding errors".




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Both!

Slap you? But why?!? I can't. I'm too gentle and soft. And you're too harmless. Well, maybe a little. *soft slap*... *little less of a soft slap*...

*slaps!*
*SLAPS!!*
*THUMPS!!!!*
*MONSTER THUMP!*

Woah, sorry, lost control

Told you I'm a fast learner and adapter. ^^

I like Serena

Yes dear madam, it is too much for rounding errors. ;)

How did you get the angle of 28.5 degrees?


Aha, now I understand the head band.
It is both to protect your forehead from the slap, and to hide the bruises under it!


aw, aw, aw, aw, Aw, Aw, AW, AWW, AWWWW, AIEEEH!

More please

Femme_physics

Law of sines.

BP/sin(beta) = CP/sin(130)

When I know that CP = 641.45
And BP = 306.4

I like Serena

All right, so how did you get BP and CP?

EDIT: Did you perhaps take BP from the drawing where you switched the letters A and B around?

EDIT2: You wrote in your problem that the angle is "psi", but actually that is "phi".

[itex]\phi[/itex] or [itex]\varphi[/itex] is phi

and [itex]\psi[/itex] is psi

Femme_physics

I most certainly did not switch A and B ar...no wait...wait...I most certainly did!
Your corrections are all correct-- as always. All solved ;)

thanks!