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?

- anonymous

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- anonymous

@IrishBoy123 I see you're viewing my question. Might you be able/willing to help me?

- IrishBoy123

sure

- 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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## More answers

- 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

Yes, that's correct.

- IrishBoy123

you say \(v_r = 16\)
how did you get that?
because i see the need to resolve the components along \(\vec r\)

- 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

ok
let me crunch some numbers then

- 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

if you resolve \(\vec{\dot r}\) along \(\vec r\), you get \(v_r = 16.000735\)
is that closer to their answer?

- 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

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

OK
is my answer closer to the website's?
we can deal with the rest in a minute

- IrishBoy123

|dw:1443088796660:dw|

- anonymous

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

- IrishBoy123

what they wany when they say radial velocity is this
|dw:1443088879804:dw|

- PatrickWeeWoos

Good morning all you amazing people!!!!!

- 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

Yeah, I see it.

- 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

so are we ready to do it?

- IrishBoy123

|dw:1443089158375:dw|

- anonymous

Sure. It's not what I was asking for, but sure, let's do this.

- 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

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

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

OKay.

- 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

does that make sense to you?

- anonymous

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

- anonymous

*had

- 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

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

|dw:1443091705155:dw|

- 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

i gave you a medal!

- PatrickWeeWoos

@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

thanks Pat

- anonymous

So I don't even use v at all except for when I'm solving for it?

- IrishBoy123

not for resolving acceleration into its components, no

- anonymous

But I thought acceleration was by definition, the derivative of velocity.

- anonymous

Also, it's worth mentioning that I'm not given the magnitude of the acceleration.

- anonymous

So the dot product you describe wouldn't actually work in this case.

- 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

\(\frac{32}{\sqrt{17}}\) and \(\frac{8}{\sqrt{17}}\)

- 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

Thanks, by the way, for you patience in helping me out.

- 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

Well, thanks again. I'm going to close this thread now.

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