Dynamics Problem, Relative velocity of rigid body in planar motion

[CaityAnn]

http://wps.prenhall.com/wps/media/objects/3076/3149958/studypak/questions/html/Ch16/9e_16_43.html
That is the image of the problem, which includes a solution. PROBLEM- Besides being really confused on their work, the solution they give and the solution in the back of my book are both different! They said Wcd is 15.1 rad/s and my book says 4.03 rad/s, they have the same dimensions and everything.


heres what I am doing, despire the crazy solution given.. I am just trying to get some sort of method and understanding of this stuff, I am taking this course in 5 weeks so its sort of a rush to get it all in..

I want to find Angular velocity of link CD at the instant shown

Vc=Vd+Vc/d ; Vc=Vd+(Wcd x Rcd) Vd won't be moving so it goes to zero
So Vc=(WcdxRcd) ((((So I must need to solve for Vc to get Wcd))))

Va= Vb+ Va/b; Va= Vb + (Wab x Rab), Va won't be moving so it goes to zero
Vb= (Wabx Rab), I can solve for this because I am given Wab and Rab, Vb=18 in/s

Vc=Vb+Vc/d; Vc=Vb+(WcdxRcd) Vc= 8 + (WcdxRcd)
So I just need to find Vc somehow.. On the solution given they do some insane trig that I just don't get, your input is much appreciated so thanks ahead..

 

[AlephZero]

I agree with your book's solution.

First suppose D is free to move, and write down the motion of points B C and D in terms of the angular velocities (positive anticlockwise)

Vb = (0, -18) in/sec
Vc = Vb + (8Wbc cos 30, -8 Wbc sin 30)
Vd = Vc + (4Wcd sin 45, 4 Wcd cos 45)

But Vd = 0 since it is pinned. So

8 Wbc cos 30 + 4 Wcd sin 45 = 0
-18 - 8 Wbc sin 30 + 4 Wcd cos 45 = 0

Eliminating Wcd (remembering sin 45 = cos 45) gives

-18 - 8 Wbc (sin 30 + cos 30) = 0
Wbc = -1.647 rad/s

And
Wcd = -2 Wbc cos 30/sin 45 = 4.034 rad/s

I didn't try to find what why this is different from the web page solution.