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?

Question

anonymous · Answer

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

irishboy123 · Answer

sure

anonymous · Answer

OKay, cool. So like I said, I can't figure out what the question is trying to get me to do.

irishboy123 · Answer

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 · Answer

Yes, that's correct.

irishboy123 · Answer

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

anonymous · Answer

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 · Answer

ok

let me crunch some numbers then

irishboy123 · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

irishboy123 · Answer

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anonymous · Answer

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

irishboy123 · Answer

what they wany when they say radial velocity is this
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