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Thread: Vector kinematics - terms vanishing?

  1. #1 Vector kinematics - terms vanishing? 
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    Hello everyone,
    This is my first post here - I look forward to looking around to see what goes on here.

    Anyway, I'm a mechanical engineering senior, taking a vector kinematics class. And I've been stumped all evening. I suspect I'll feel dumb when somebody points out why this is occuring (hoping it's just due to sleep deprivation lol), but I can't for the life of me figure out how the solution I am looking at is doing away with a pair of terms.

    I'd like to think I have a pretty decent grasp on this material, but the 4th picture I posted (whole problem is spread over 4 photos) is throwing me for a loop. I'm good with it until they say "Simplifying" and drop two of the r_2 velocity terms. I've looked at a number of answers in the solutions I found online and they all seem to drop a term at the end for no apparent reason. My textbook is still on backorder and google hasn't turned up much that's relevant. I have literally spent the last 6 hours puzzling over this, so I will be tremendously grateful to anyone who can shed some light on it. Why do the r_2'Cos(theta_2) and r_2'sin(theta_2) terms just vanish without warning during the "simplifying" phase?

    https://www.facebook.com/photo.php?f...9&l=4026c64019

    https://www.facebook.com/photo.php?f...8&l=140208c71d

    https://www.facebook.com/photo.php?f...9&l=fa3ca6dbd2

    https://www.facebook.com/photo.php?f...1&l=c6c539a698

    Thanks.

    -Patrick
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  2. #2  
    x0x
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    Quote Originally Posted by PatrickOB View Post
    Hello everyone,
    This is my first post here - I look forward to looking around to see what goes on here.

    Anyway, I'm a mechanical engineering senior, taking a vector kinematics class. And I've been stumped all evening. I suspect I'll feel dumb when somebody points out why this is occuring (hoping it's just due to sleep deprivation lol), but I can't for the life of me figure out how the solution I am looking at is doing away with a pair of terms.

    I'd like to think I have a pretty decent grasp on this material, but the 4th picture I posted (whole problem is spread over 4 photos) is throwing me for a loop. I'm good with it until they say "Simplifying" and drop two of the r_2 velocity terms. I've looked at a number of answers in the solutions I found online and they all seem to drop a term at the end for no apparent reason. My textbook is still on backorder and google hasn't turned up much that's relevant. I have literally spent the last 6 hours puzzling over this, so I will be tremendously grateful to anyone who can shed some light on it. Why do the r_2'Cos(theta_2) and r_2'sin(theta_2) terms just vanish without warning during the "simplifying" phase?

    https://www.facebook.com/photo.php?f...9&l=4026c64019

    https://www.facebook.com/photo.php?f...8&l=140208c71d

    https://www.facebook.com/photo.php?f...9&l=fa3ca6dbd2

    https://www.facebook.com/photo.php?f...1&l=c6c539a698

    Thanks.

    -Patrick
    No image comes up.
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  3. #3  
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    Hello, welcome to the forum, the links do not work. You could try posting the images on photosharing site.
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  4. #4  
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    Quote Originally Posted by Jilan View Post
    Hello, welcome to the forum, the links do not work. You could try posting the images on photosharing site.
    My apologies - I needed to change the security settings for the facebook links. They should work now!
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  5. #5  
    x0x
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    Quote Originally Posted by PatrickOB View Post
    My apologies - I needed to change the security settings for the facebook links. They should work now!
    It is difficult to see what the equations before "Simplify" are but I am guessing that they look like this:



    But so
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  6. #6  
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    Quote Originally Posted by PatrickOB
    Hello everyone,
    This is my first post here - I look forward to looking around to see what goes on here.
    Welcome to the forum!

    Quote Originally Posted by PatrickOB
    I'd like to think I have a pretty decent grasp on this material, but the 4th picture I posted (whole problem is spread over 4 photos) is throwing me for a loop. I'm good with it until they say "Simplifying" and drop two of the r_2 velocity terms. I've looked at a number of answers in the solutions I found online and they all seem to drop a term at the end for no apparent reason.
    r2 is the magnitude of the vector r2. Notice that while r1 and r3 both change magnitude and direction r2 remains fixed while r2 changes direction. Therefore dr2/dt = 0.

    Don't you just hate it when that happens, i.e. when you look at a problem for so long and it's right there staring you in the face?

    Please take the following advice which will come in handy during your career. When you run into a problem that you're stumped on, put it down totally and literally and walk away from it. Put it out of your mind and forget about it. It will either come to you, often at an odd time and place, or when you go back to it you'll see it with fresh eyes. This is a well-known fact that all problem solvers such as physicists are very familiar with. I myself find that solutions often come to me in the "family reading room" (f you know what I mean).
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  7. #7  
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    Quote Originally Posted by x0x View Post
    It is difficult to see what the equations before "Simplify" are but I am guessing that they look like this:



    But so
    That is the correct equation. But why does dr2/dt = 0? The velocity of vector 2 describes the motion of point 3, correct? And point 3 has motion in both the y direction (sliding vertically in joint 4) and horizontal motion as link 4 slides horizontally.

    Thank you,
    Patrick
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  8. #8  
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    Quote Originally Posted by PatrickOB View Post
    That is the correct equation. But why does dr2/dt = 0?
    That's part of the problem. Look at the problem statement and you'll see that it says
    Also, length O2A = 20 inches.
    That means that the rod O2A has a fixed length of 20 inches. See this diagram Scotch yoke - Wikipedia, the free encyclopedia and find that rod and you'll see what it means for it to remain a constant length.
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  9. #9  
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    Ooooh. Wow, I'm an idiot. So the velocity of the vector describes the rate of change of the magnitude of the vector, not the point in space at which it ends. Which should've been obvious, since it's the derivative of the magnitude (and the d*theta/d*t describes the angular change). I was looking at it all wrong.

    Thanks guys!
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  10. #10  
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    Quote Originally Posted by PatrickOB View Post
    Ooooh. Wow, I'm an idiot. So the velocity of the vector describes the rate of change of the magnitude of the vector, not the point in space at which it ends. Which should've been obvious, since it's the derivative of the magnitude (and the d*theta/d*t describes the angular change). I was looking at it all wrong.

    Thanks guys!
    You are welcome!
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  11. #11  
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    Quote Originally Posted by PatrickOB View Post
    Ooooh. Wow, I'm an idiot. So the velocity of the vector describes the rate of change of the magnitude of the vector, not the point in space at which it ends. Which should've been obvious, since it's the derivative of the magnitude (and the d*theta/d*t describes the angular change). I was looking at it all wrong.

    Thanks guys!
    No. d|r|2/dt is a scalar where as r2 is a vector.

    d|r|2/dt is called speed and dr2/dt is called velocity. Speed is the magnitude of velocity.

    Note- I was informed that x0x made the comment that "No Pete dr2/dt is velocity." and then made some rude comments about me confusing people. Instead of posting snide comments like this bruce, why not just be more useful instead and point out any typos you might see. You might be useful for spotting typos. Wasn't that part of your last job?

    -------------------
    x0x made an error below. Instead of letting me know that I forgot to add the differential operator d/dt to r2 he instead made false accusations regarding the reason for why I forgot it. He also made the silly comment "is just the positional vector, you are confusing people with your mistakes." which is a terribly insulting thing to say. The content of all my posts in this thread explains that brucep. And all the decent people here know that my math skills are light years ahead of yours so please don't act like i don't know what a position vector is since that's what I was talking about in detail. I'm far too intelligent and highly educated to confuse the position vector with speed. Especially in this thread since that's what I was explaining to the OP. Lol!!! To think I actually confused the two is the real laugh!!! ROTFL!!! That's for the good laugh, bruce!!!
    ---------------------

    When will this forum stop members from posting comments like that?
    Last edited by Physicist; 09-21-2014 at 12:27 PM.
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  12. #12  
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    Quote Originally Posted by Physicist View Post
    r2 is called velocity.
    No, Peter, velocity is .

    is just the positional vector, you are confusing people with your mistakes.
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