anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@IrishBoy123 I see you're viewing my question. Might you be able/willing to help me?
IrishBoy123
  • IrishBoy123
sure
anonymous
  • anonymous
OKay, cool. So like I said, I can't figure out what the question is trying to get me to do.

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IrishBoy123
  • IrishBoy123
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?
anonymous
  • anonymous
Yes, that's correct.
IrishBoy123
  • IrishBoy123
you say \(v_r = 16\) how did you get that? because i see the need to resolve the components along \(\vec r\)
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
ok let me crunch some numbers then
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
if you resolve \(\vec{\dot r}\) along \(\vec r\), you get \(v_r = 16.000735\) is that closer to their answer?
anonymous
  • anonymous
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).
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
OK is my answer closer to the website's? we can deal with the rest in a minute
IrishBoy123
  • IrishBoy123
|dw:1443088796660:dw|
anonymous
  • anonymous
Yes. to three sig figs, it is \[v _{r} = 16.0\]. But like I said, we already knew that one.
IrishBoy123
  • IrishBoy123
what they wany when they say radial velocity is this |dw:1443088879804:dw|
PatrickWeeWoos
  • PatrickWeeWoos
Good morning all you amazing people!!!!!
IrishBoy123
  • IrishBoy123
do you see it yet v is the tangential velocity of the paticle \(\vec v_r\) is the velocity resolved along \(\vec r\)
anonymous
  • anonymous
Yeah, I see it.
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
so are we ready to do it?
IrishBoy123
  • IrishBoy123
|dw:1443089158375:dw|
anonymous
  • anonymous
Sure. It's not what I was asking for, but sure, let's do this.
IrishBoy123
  • IrishBoy123
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
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
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|
anonymous
  • anonymous
OKay.
IrishBoy123
  • IrishBoy123
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 \(\large v_t = \vec {\dot r} \bullet \hat{ t} = <2,16>\bullet \frac{<-4,1>}{\sqrt{17}} = 1.94m/s\)
IrishBoy123
  • IrishBoy123
does that make sense to you?
anonymous
  • anonymous
You have me up until you have \[t = \frac{ <-4, 1> }{ \sqrt{17} }\]. Where'd you get that?
anonymous
  • anonymous
*had
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
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 \( \vec{\dot r}\).
IrishBoy123
  • IrishBoy123
|dw:1443091705155:dw|
anonymous
  • anonymous
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?
PatrickWeeWoos
  • PatrickWeeWoos
i gave you a medal!
PatrickWeeWoos
  • PatrickWeeWoos
@IrishBoy123
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
thanks Pat
anonymous
  • anonymous
So I don't even use v at all except for when I'm solving for it?
IrishBoy123
  • IrishBoy123
not for resolving acceleration into its components, no
anonymous
  • anonymous
But I thought acceleration was by definition, the derivative of velocity.
anonymous
  • anonymous
Also, it's worth mentioning that I'm not given the magnitude of the acceleration.
anonymous
  • anonymous
So the dot product you describe wouldn't actually work in this case.
IrishBoy123
  • IrishBoy123
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 \(\vec a_t = \vec a \bullet \hat t\)
IrishBoy123
  • IrishBoy123
\(\frac{32}{\sqrt{17}}\) and \(\frac{8}{\sqrt{17}}\)
anonymous
  • anonymous
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...
anonymous
  • anonymous
Thanks, by the way, for you patience in helping me out.
IrishBoy123
  • IrishBoy123
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
anonymous
  • anonymous
Well, thanks again. I'm going to close this thread now.

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