Relative Velocity and angular velocity

[_Bd_]

Homework Statement

Knowing that at the instant shown rod AB has a constant angular velocity (w = 15 rad/s) counterclockwise (+k) determine (a) the angular acceleration of rod DE (b) the acceleration of point D

Homework Equations

(w = omega)

V_B = V_A + V_B/A
V_B/A=w X r
a_B = a_A + a X r + w X (wXr)

The Attempt at a Solution

I used the first formula assuming that Va=0 since its pinned, same for its acceleration. V_B/A=w X r so then Vb = 60j + 45 i and therefore aB = 900 i + 675 j
so far so good. I then started to solve for the rod B-C and then the rod C-D

the professor told me you can solve for the rod B - D instead, although I don't understand necessarily why I attempted it and got lost.

The professor posted the solution in her webpage and this is what trips me out the most:

[PLAIN]http://img194.imageshack.us/img194/4236/solutioni.jpg
http://img194.imageshack.us/img194/4236/solutioni.jpg

That part I highlighted is complteley unknown to me, why is it B/D and not D/B
Im also from what I see, using that system of equations (rod B - D and rod D - E)
it says on the j part component -4w_D/E = 60 + 10w_AB
I have no clue where did that come from. . .can someone elaborate on that whole part please?

 

[mark726]

Thank you for your post. It seems that you are on the right track with your solution, but there are a few key concepts that you seem to be struggling with. Let's break down the problem and go through it step by step.

Firstly, it is important to understand that the angular velocity of a rigid body is the same at all points on the body. This means that the angular velocity of rod AB is also the angular velocity of rod BC, CD, and DE. This is why the professor used the same symbol for the angular velocity of all the rods.

Next, let's look at the formula VB/A=w X r. This formula is used to find the velocity of point B relative to point A. In this problem, point A is the pivot point of rod AB, and point B is the end of the rod. So, using this formula, you correctly found the velocity of point B relative to point A.

Now, let's look at the next formula, aB=aA + a X r + w X (wXr). This formula is used to find the acceleration of point B relative to point A. The first term, aA, represents the acceleration of point A. Since point A is pinned, it has no acceleration, so this term is zero. The second term, a X r, represents the acceleration of point B due to the rotation of the rod. This term is zero because the rod is rotating at a constant angular velocity, so there is no change in its rotational motion. The last term, w X (wXr), represents the centripetal acceleration of point B due to its circular motion. This is the term that you are missing in your solution. If you plug in the values for w and r, you will get the answer given by the professor.

Now, let's move on to part (b) of the problem. The key concept that you seem to be struggling with is that the acceleration of a point on a rigid body is the same as the acceleration of the body itself. In other words, the acceleration of point D is the same as the acceleration of rod DE. This is why the professor used the same symbol for both in the solution.

I hope this helps to clarify some of the confusion you had with the solution. If you have any further questions, please don't hesitate to ask. Keep up the good work and happy studying!
Scient