Question about Slider Crank Mechanism problem

[subatomicprincess]

Homework Statement

For the slider-crank mechanism shown, determine V_D and omega_(BD) with crank AB rotating at a constant angular velocity of 8- rad/s CW for theta = 150 degrees.

Relevant Equations

V_B = omega_(AB) x r_(AB)
V_(D/B) = omega_(D/B) x r_(D/B)

I have attached my attempt. The only thing I keep getting wrong is the direction of V_(D/B). I know it is perpendicular to r_(BD) but, I never know if it is going upwards or downwards. In this case, I thought it was going downwards but when I checked the answer key, it had said it is going upwards. Don't know why though.

 

[haruspex]

Not sure what you mean by V_(D/B). Do you mean V_D, i.e. the velocity of D?
Yes, if AB is rotating clockwise then clearly D is moving down and to the right.

 

[subatomicprincess]

V_(D/B) is how the professor explained it but it is the velocity of rod BD. V_D is the velocity of D.

I just got confused because the vector V_D was drawn at the top of the triangle and V_(D/B) was pointing towards it (in the velocity diagram). But now I know velocity BD points toward velocity D when it is moving clockwise! Thank you!

 

[Lnewqban]

Could you post a diagram of the mechanism?
Thank you... and welcome to PF!

 

[haruspex]

V_(D/B) is how the professor explained it but it is the velocity of rod BD. V_D is the velocity of D.

I just got confused because the vector V_D was drawn at the top of the triangle and V_(D/B) was pointing towards it (in the velocity diagram).

Still not following.
What is meant by the "velocity of BD"? BD is rotating about B, yes? I thought you might have meant angular velocity, but you have a Relevant Equation showing it is the velocity at some point distance r from B. So what is r?
If V_(D/B) is the velocity of the midpoint of BD then that should be parallel to V_D.
And I don't know what it would mean for one vector to point towards another vector:

But now I know velocity BD points toward velocity D when it is moving clockwise! Thank you!

 

[Master1022]

Hi there,

I don't agree with your velocity triangle (I think - it isn't too clear). Specifically the fact that $V_{BD}$ doesn't start at the end of $v_B$. Apologies if this is understood, but I understand $v_{BD}$ to be the "velocity of D relative to B" and thus the vector $v_{BD}$ should start at the end of $v_B$. (EDIT: maybe the convention I have learned is back-to-front, but either way the velocity should not start at the same point where B starts)

The way I would attempt this problem is:
1. Draw $v_B$ (which you have done)
2. Draw a line that indicates the range of possible motion of D.
What does that mean?
If D is constrained to move horizontally, then you know that $v_D$ must lie in the horizontal plane, so you can just draw a dotted line extending in both directions to denote that we expect D to be somewhere along that line. (Same idea can be applied for any type of directional constraint)

Spoiler: Velocity diagram sketch

So your diagram will look something like. I have purposely left out the angles.

3. We know B and we want to find where $D$ is - we need another vector ($v_{BD}$) to fix this point. So now let us imagine we are rotating the crank in the CW direction. If we are stood at B, then which direction will D appear to be rotating (relative to B)?? If you aren't sure, then you could do another dotted line to allow for rotation in either direction. Then draw this vector/line starting from B. Now you should have an intersection between the two lines along which D can lie and thus you can find D.

Spoiler: Velocity diagram sketch pt. 2

Echoing sentiments from @haruspex and @Lnewqban above, it would be very helpful if you can post a picture of the actual physical mechanism. Just so I don't misinterpret the actual situation.

Hope this was of some help. Some details with drawing the velocity triangle have been left out as it is somewhat tedious to explain it all in a post if you already understand it.