# Velocity of linkage arm end

1. Apr 5, 2017

### Dusty912

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

I uploaded the picture to this problem, but it asks what the relative velocity of point p is to point o when theta is equal to one and the derivative of theta in respect to time is also positive 1. the picture explains everything.

2. Relevant equations

Tangential velocity is equal to omega (angular velocity) times the radius. the radius would be the 200 mm arm in this case.

3. The attempt at a solution

Honestly, my solution seems too simple, but the logic seems right to me. I basically just used the equation
Vt=ω*r
the Vt=200*1=200

then just used the cosΘ=Vxt/Vt

and solved for Vxt with Θ=1 rad and Vt = 200

the answer I got was 108.0598mm/s

I went off the idea that the Vtx value would always have to be equal to the velocity of P

I may have gone horribly wrong somewhere but I'm not really sure if I have. Thanks for the help, you guys and girls rock!!

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Last edited by a moderator: Apr 6, 2017
2. Apr 6, 2017

### andrevdh

I don't think your assumption is correct.
The motion of P would only be the same if the x distance between the two point were fixed, which it is not, it changes as the connection point moves up or down.

3. Apr 6, 2017

### TomHart

This is a tricky little problem that I spent quite a bit of time thinking about.
Yeah, I am in agreement with andrevdh that your assumption is not correct. Here's why I think that. Start by disconnecting the two rods at the connection point (I am going to call that point T for Top) and taking the shorter rod completely out of the picture. Then if you pull vertically upward at point T on the remaining rod (the right-hand rod), that will produce movement of point P to the left. Note that leftward velocity of point P can be produced by vertical velocity (only) of point T. For that reason, your assumption is incorrect that the velocity of point P is equal to the x component of the velocity of point T. The only times when the velocity of P is equal to the horizontal velocity of point T is when the left rod is vertical (θ = 90°) - the point when the vertical component of the velocity of point T is equal to 0, or when the left rod is horizontal (θ = 0°) - the point when the horizontal component of the velocity of point T is equal to 0. Edit: I reversed θ = 0° and θ = 90° above; I had them backwards.

The way I see it, the velocity of point P in the x direction is the sum of two components:
1) the velocity of point P due to the x component of the velocity of point T (these two are equal)
2) the velocity of point P due to the y component of the velocity of point T (the relative size being dependent on the angle θ)

So I agree that your answer of 108.0598 mm/s is the x component of the velocity of point T. Now you have to find the velocity of point P produced by the vertical component of the velocity of point T.

Please post when you come up with a final answer so we can compare results. Thank you.

Last edited: Apr 6, 2017
4. Apr 6, 2017

### Dusty912

sp would I just have to include the vertical component of the tangential velocity as well? not really sure where to go from here

5. Apr 6, 2017

### TomHart

You can't just include the vertical component of the tangential velocity; you have to calculate the horizontal velocity of point P due to the vertical velocity of [what I call] point T.
Once again, imagine that the left rod has been removed and that the right rod is lying horizontally. If you grab ahold of point T and start to pull it vertically upward with constant velocity, point P will start moving very slowly, at first, toward the left. But the higher point T gets, the greater the velocity that point P will move toward the left. The ratio of those two velocities (point T's upward velocity relative to point P's leftward velocity) changes as the angle θ changes.

6. Apr 6, 2017

### Nidum

Clue 1 : This is in the problem statement . Why do they quote the angular velocity in differential form ?

Clue 2 : Write down a simple geometric function relating angle of crank to axial position of crosshead .

Last edited: Apr 6, 2017
7. Apr 6, 2017

### TomHart

Thank you for your comment. I think this method is more straightforward than what I was suggesting - once one can actually come up with that geometric function. Just to clarify due to my lack of understanding of the terms, "axial position of crosshead" is basically point P's position along the x axis. Is that correct? Thank you.

8. Apr 6, 2017

### Nidum

Yes - that's it .

9. Apr 6, 2017

### Dusty912

well the law of cosines relates all of those

10. Apr 6, 2017

### Nidum

Use a construction line dropping vertically from crank pin to horizontal axis .

11. Apr 6, 2017

### Dusty912

so then tanΘ=(opposite)/200mm

12. Apr 6, 2017

### Dusty912

sorry not tangent

13. Apr 6, 2017

### Dusty912

I meant sine or cosine relations could be used

14. Apr 6, 2017

### Dusty912

15. Apr 6, 2017

### TomHart

Yes, keep going. I am going to take a back seat on this one and defer to Nidum.

16. Apr 6, 2017

### Dusty912

hmm could you just let me know if this next step is right? So this would be my position function. by taking the derivative with respect to time, I would get my velocity function?

17. Apr 6, 2017

### Nidum

Yes- that's it .

But you haven't got the geometric function yet .

18. Apr 6, 2017

### Dusty912

ok but wouldn't the derivative of say the cosine function be -sinΘ=0 (my velocity function)

19. Apr 6, 2017

### Nidum

Don't quite understand what you are saying there ?

20. Apr 6, 2017

### Dusty912

well I said earlier that cosΘ=(adjacent)/200mm would be my position function. and that it's derivative with respect to time would be my velocity function? I guess I made a mistake in my derivation. But how should I approach this derivation?