What is the Velocity of Point P relative to Point O in a Linkage Arm System?

[haruspex]

It's saying that THE PROJECTIONS of V_p and V_t are changing at the same rate

Vp and Vt are velocities. There is a simpler relationship than just that they change at the same rate.

 

[Dusty912]

they are equal to each other

 

[Dusty912]

well actually that doesn't really make any sense, since V_t is a constant velocity and V_p is not

 

[Dusty912]

is it just that the projections are always equal to each other?

 

[haruspex]

well actually that doesn't really make any sense, since V_t is a constant velocity and V_p is not

The constancy does not matter since we are only concerned with instantaneous velocities at the given position. Anyway, even if |Vt| is constant the vector Vt is not..

is it just that the projections are always equal to each other?

Yes. Do you see why?

 

[Dusty912]

yes I see why. because the vector is dependent upon magnitude and direction

 

[Dusty912]

ok so the horizontal components are always equal to each other. but don't the vertical compontents of the tangential velocity also effect the horizontal velocity at P?

 

[haruspex]

ok so the horizontal components are always equal to each other.

No, as you wrote earlier:

the projections are always equal to each other?

.. but projections in which direction? What direction did I say to consider?

because the vector is dependent upon magnitude and direction

No. What physically constrains the relative motions of P and Q?

 

[Dusty912]

the projections in the horizontal direction are the ones we are considering?

the 400mm are physically constrains these two

 

[haruspex]

the projections in the horizontal direction are the ones we are considering?

No.

the 400mm are physically constrains these two

Right. So what does that mean for their relative motion in the PQ direction?
You might find it easier to think about in the frame of reference of P. Relative to P, can Q get any closer or further away? What does that say about the direction of the relative velocity of Q?

 

[Dusty912]

no Q cannot get any closer to P, the velocities would be in the same direction for P and Q

 

[Dusty912]

ok so I attempted the problem again with the information you gave me. I basically found the angle between V_t and the action line. which was 32.415 degrees. Then I found V_t by multiplying the radius by the angular velocity (200mm/s) the I used cos(32.415)*200mm/s to find my answer. so basically I used the graphic that the other member had drawn for me. and projected V_t onto the line PQ

 

[Dusty912]

so the velocity of P should match the velocity of V_t onto the line PQ. my logic was because when the rotating arm is at 0 radian, The velocity at P is zero, and so is the velocity of V_t projected onto the line.

 

[haruspex]

so the velocity of P should match the velocity of Vt onto the line PQ.

The components of the velocities in the PQ direction should be the same, yes. What answer does that give you?

when the rotating arm is at 0 radian, The velocity at P is zero, and so is the velocity of Vt projected onto the line.

Yes, and that fits with the general relationship described above. Because PQ is a constant distance, the relative velcities of P and Q must always be the same in the PQ direction.

 

[Dusty912]

the answer I got was 168.84 mm/s

 

[Dusty912]

and is the max speed of P, the tangential speed of the arm (200mm/s)? it couldn't go faster than that right? just curious about that

 

[TomHart]

I got an answer greater than 200 mm/s for this problem: However, I have very low confidence in my result. Is there anyone else who has worked this problem through and can comment on Dusty's result of 168.84 mm/s - whether it is correct or not?

Edit: By the way, I stumbled in trying to find a relationship between the angle θ and the position of point P - using law of cosines or any other method - along the x axis. I was not able to do it.

 

[gneill]

I got an answer greater than 200 mm/s for this problem: However, I have very low confidence in my result. Is there anyone else who has worked this problem through and can comment on Dusty's result of 168.84 mm/s - whether it is correct or not?

Edit: By the way, I stumbled in trying to find a relationship between the angle θ and the position of point P - using law of cosines or any other method - along the x axis. I was not able to do it.

I also find a value greater than 200 mm/s.

If you use the law of cosines then you'll have to solve a quadratic and choose the correct root. Another approach is to break the geometry into a pair of triangles and use a mix of trig and Pythagoras. Both methods yield the same expression for the length OP in terms of θ.

 

[Dusty912]

I am going to try it with law of cosines. it looks like it will make more sense that way.