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.

 

[ravenprp]

It seems like you are on the right track with your approach. The key concept in this problem is the relative velocity of a rigid body in planar motion. This means that the velocity of one point on the body can be determined by the velocity of another point on the body, as long as the relative position and angular velocity of these points are known.

In this case, we are given the velocity of point B (Vb) and the angular velocity of link AB (Wab). Using the equation Vb = (Wab x Rab), we can solve for Vb.

Next, we can use the equation Vc = Vb + (Wcd x Rcd), where Vc is the velocity of point C and Vd is the velocity of point D. Since Vd is zero, we can simplify this equation to Vc = Vb + (Wcd x Rcd).

Now, we need to find Vc in order to solve for Wcd. To do this, we can use the equation Vc = (Wcd x Rcd), where Vc is the velocity of point C. This equation is based on the fact that the velocity of a point on a rotating body is equal to the angular velocity of the body multiplied by the distance from the axis of rotation.

So, by substituting Vc = Vb + (Wcd x Rcd) into Vc = (Wcd x Rcd), we can solve for Wcd. This will give us the angular velocity of link CD at the given instant.

It is possible that the solution given in your book or the one provided in the link may have made a mistake in their calculations. It's always a good idea to double check your work and try to understand the concepts behind the equations instead of just relying on the solution provided. Keep up the good work and don't get discouraged! With practice and understanding, you will be able to tackle more complex dynamics problems.