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anonymous

  • one year ago

Hello, I'm having a little trouble with this one problem I'm on. It gives me vector r = {2t i + 4t^2 j} ft and it asks me to determine the radial component and the transverse component of the particle's velocity at t = 2 s. I've already found the radial component of the velocity, v_r = 16.0 ft/s. But I can't figure out how to find the transverse component. I know v_theta = r(deriv. of theta), but I can't figure out where I'm supposed to get theta from. I've tried a couple of things, but neither of them worked. Is there anyone online that can help me?

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  1. anonymous
    • one year ago
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    @IrishBoy123 I see you're viewing my question. Might you be able/willing to help me?

  2. IrishBoy123
    • one year ago
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    sure

  3. anonymous
    • one year ago
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    OKay, cool. So like I said, I can't figure out what the question is trying to get me to do.

  4. IrishBoy123
    • one year ago
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    first of all we are working along the vector \(\vec r = <2t, 4t^2> = <4,16>_{t=2}\) which gives us \( \vec {\dot r} = <2,8t> = <2,16>_{t=2}\) is that agreed?

  5. anonymous
    • one year ago
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    Yes, that's correct.

  6. IrishBoy123
    • one year ago
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    you say \(v_r = 16\) how did you get that? because i see the need to resolve the components along \(\vec r\)

  7. anonymous
    • one year ago
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    That would be the magnitude of v_r, which is roughly 16.0 ft/s. MasteringEngineering said I was right, so I believe it is so. Although Mastering's answer is rounded a little different, as the actual answer I got was closer to 16.12 ft/s.

  8. IrishBoy123
    • one year ago
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    ok let me crunch some numbers then

  9. IrishBoy123
    • one year ago
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    mmm what you have calculated is \(|\vec{\dot r}| = \sqrt{2^2 + 16^2} = 16.12\) ie you have calculated the speed of the particle

  10. IrishBoy123
    • one year ago
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    if you resolve \(\vec{\dot r}\) along \(\vec r\), you get \(v_r = 16.000735\) is that closer to their answer?

  11. anonymous
    • one year ago
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    Yes. I found \[v _{r}\]. I am looking for \[v _{\theta}\], which is described as v_theta = r(dtheta/dt) (I don't know how to write it with the dot overhead on this thing).

  12. anonymous
    • one year ago
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    My problem is I don't know how to get theta, or if it's even relevant here, since everything seems to be in cartesian coordinates.

  13. IrishBoy123
    • one year ago
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    OK is my answer closer to the website's? we can deal with the rest in a minute

  14. IrishBoy123
    • one year ago
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    |dw:1443088796660:dw|

  15. anonymous
    • one year ago
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    Yes. to three sig figs, it is \[v _{r} = 16.0\]. But like I said, we already knew that one.

  16. IrishBoy123
    • one year ago
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    what they wany when they say radial velocity is this |dw:1443088879804:dw|

  17. PatrickWeeWoos
    • one year ago
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    Good morning all you amazing people!!!!!

  18. IrishBoy123
    • one year ago
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    do you see it yet v is the tangential velocity of the paticle \(\vec v_r\) is the velocity resolved along \(\vec r\)

  19. anonymous
    • one year ago
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    Yeah, I see it.

  20. IrishBoy123
    • one year ago
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    to get that \(\vec v_r\), you need to find the unit vector along \(\vec r\), and then you claculate the dot product of that and \(\vec v\), that resolves \(\vec v \) radially

  21. IrishBoy123
    • one year ago
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    so are we ready to do it?

  22. IrishBoy123
    • one year ago
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    |dw:1443089158375:dw|

  23. anonymous
    • one year ago
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    Sure. It's not what I was asking for, but sure, let's do this.

  24. IrishBoy123
    • one year ago
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    your answer is out because you have calculated the "speed" of the particle ie \(\sqrt{2^2 + 16^2}\) that is not the radial component. the radial component is gotten this way do you wish to proceed? i honestly do not want to waste your time if this is not what you want to do just let me know

  25. anonymous
    • one year ago
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    What's the difference? Like I said, I used the radial speed for the problem I'm doing right now and the site already said that I was right. I'm looking for transverse velocity at this point.

  26. IrishBoy123
    • one year ago
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    well the site is wrong as there is a rounding error in there. i thought we had agreed that! we can skip the radial bit and go to the tangential bit then we need the unit tangent vector |dw:1443089994080:dw| so that we can resolve the particles velocity in that direction |dw:1443090134203:dw|

  27. anonymous
    • one year ago
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    OKay.

  28. IrishBoy123
    • one year ago
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    but we need to the unit radial to get that we have \(\large \vec r = <2t, 4t^2> = <4,16>_{t=2}\) so \(\large \hat{ r} = \frac{<4,16>}{\sqrt{4^2+16^2}} = \frac{<1,4>}{\sqrt{17}}\) so \(\large \hat{ t} = \frac{<-4,1>}{\sqrt{17}}\) so \(\large v_t = \vec {\dot r} \bullet \hat{ t} = <2,16>\bullet \frac{<-4,1>}{\sqrt{17}} = 1.94m/s\)

  29. IrishBoy123
    • one year ago
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    does that make sense to you?

  30. anonymous
    • one year ago
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    You have me up until you have \[t = \frac{ <-4, 1> }{ \sqrt{17} }\]. Where'd you get that?

  31. anonymous
    • one year ago
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    *had

  32. IrishBoy123
    • one year ago
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    it comes directly from from \(\hat r\): switch the x,y and multiply one of them by -1 and you get the perpendicular vector. a bit like \(m_1 * m_2 = -1\) if 2 straight lines are perpendicular. try it: \(\hat r \bullet \hat t = 0\) also, in terms of radial velocity: \[\large v_r = \vec {\dot r} \bullet \hat{ r} = <2,16>\bullet \frac{<1,4>}{\sqrt{17}} = 16.007m/s\] which is the correct radial velocity

  33. IrishBoy123
    • one year ago
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    Oh, a word about the word "transverse" i have been using the term tangential . just to be totally clear, the transverse is what we are calculating and have calculated, and is the one that runs perpendicular to the radial as per the drawings clearly there is a tangent to the path of the particle too but that is just \( \vec{\dot r}\).

  34. IrishBoy123
    • one year ago
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    |dw:1443091705155:dw|

  35. anonymous
    • one year ago
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    Okay, I see what you're saying. Going by that logic then, let's say I wanted to find the components for acceleration next. Going by the steps you've presented, I would basically be repeating the same process over again, right? Only with the velocity vector in the place of the position vector. Resolve a_t and a_n both tangentially and in the direction of v_r, respectively, and then find t to solve for each dot product? Or am I missing something?

  36. PatrickWeeWoos
    • one year ago
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    i gave you a medal!

  37. PatrickWeeWoos
    • one year ago
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    @IrishBoy123

  38. IrishBoy123
    • one year ago
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    stick with the \(\hat r\) and \(\hat t\) for resolving purposes and calculate \(\vec a \bullet \hat r\) and \(\vec a \bullet \hat t\) for radial and transverse purposes

  39. IrishBoy123
    • one year ago
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    thanks Pat

  40. anonymous
    • one year ago
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    So I don't even use v at all except for when I'm solving for it?

  41. IrishBoy123
    • one year ago
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    not for resolving acceleration into its components, no

  42. anonymous
    • one year ago
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    But I thought acceleration was by definition, the derivative of velocity.

  43. anonymous
    • one year ago
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    Also, it's worth mentioning that I'm not given the magnitude of the acceleration.

  44. anonymous
    • one year ago
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    So the dot product you describe wouldn't actually work in this case.

  45. IrishBoy123
    • one year ago
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    oh! ok, crossed wires here we have \[\vec {\dot r} = <2,8t> = <2,16>_{t=2}\] whence \(\vec a = \vec {\ddot r} = <0,8> \) and \(\vec a_r = \vec a \bullet \hat r\), and \(\vec a_t = \vec a \bullet \hat t\)

  46. IrishBoy123
    • one year ago
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    \(\frac{32}{\sqrt{17}}\) and \(\frac{8}{\sqrt{17}}\)

  47. anonymous
    • one year ago
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    Just ran the numbers. You were right. Damn, if what my Dynamics prof. taught in the lecture turn out to be totally useless. Equations did not apply here; needed visual representations and dot products to get the picture. This is embarrassing...

  48. anonymous
    • one year ago
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    Thanks, by the way, for you patience in helping me out.

  49. IrishBoy123
    • one year ago
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    you can do this with loads and loads of fiddling around too, resolving things here and there. this is just an easy way to get there. mp:p

  50. anonymous
    • one year ago
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    Well, thanks again. I'm going to close this thread now.

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