Question

Hey :) I need to explain what the Relationship between radial distance and tangential velocity is (Graphed). Ive tried to figure it out but i just don't get it.

Answer 1

Hi! First, do you know what both tangential velocity and radial distance is?

Answer 2

Also, is this physics with algebra, or calculus?

Answer 3

This is Physics :)
and no honestly i have no idea o.O i don't even know how i mad it this far in physics

Answer 4

Haha! I got ya. Sometimes it's really hard to keep up with the new stuff.

So, we are talking about circular motion! Like, a ball swinging around on a string! The radial distance is the distance from the ball to the point of rotation (which is like your hand). You can see how it got it's name - it's the radius of the ball's circular path!

Is that part okay?

Answer 5

Yea that seems simple enough :P

Answer 6

Yeah!

Tangential velocity is a little different. It is what you would think of as the velocity! The only reason we give it a special name is that there is another thing called angular velocity. For tangential, it's the velocity it has, and the velocity you would see if it was let go.

The name comes from the word "tangent." That's like parallel to the motion. Picture!

Answer 7

The velocity is distance per time, right?

We say that the time of the rotation is the "period," \(T\).
The distance is the circumference, \(2\pi r\).

So distance per time is \(\dfrac{2\pi r}{T}\).

That is, \(v=\dfrac{2\pi r}{T}\).

What is the relationship, then? Well, I think it depends on what you have graphed, right?

I have to go, but I'll look at this post later! :) Good luck!

Answer 8

Thanks Sooooo much. U already explained more then my teacher :*

Answer 9

Hello again! Thanks for the image. Now, the question asked for just the relationship between radial distance and tangential velocity? Is that what the question askes specifically?

Here's a bunch of facts on a good site to refer to. It reminded me of another common way to talk about tangential velocity. It's another formula that we don't need to go into if you don't want.
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

Answer 10

Unrelated, the graph looks similar to an exponentially decreasing function.. Really, the tangential velocity is decreasing as the radial distance increases.

Answer 11

"Using your knowledge of graphs, describe the relationship between radial distance and tangential velocity. "
this is the exact question.
my problem is that, 1) i dont get what they mean by "relationship"
and 2) i wouldn't know how to enplane it o.O

Answer 12

Anybody else have input? :)

Answer 13

I'll add input later if needed... i teach physics but i'm busy atm

Answer 14

Thanks! :)

Answer 15

As for the relationship, part...
Everything in math is relationships between numbers, or amounts! :)

So, you have velocity \(v\) and radius \(r\), and we know that \(v=\dfrac{2\pi r}{T}\). That's a specific relationship. Others are like, \(1 < 2\), \(4=4\), etc...

For this problem, I'm not sure if the math is necessary, since it says knowledge of graphs. But if you want to get into it anyway, you can read this paragraph. Really, \(v\) is "directly proportional" to \(r\). They are related proportionally, so if you increase \(r\), \(v\) increases by a fraction of how much \(r\) changed. And the "direct" part is why we say it increases. So, \(v=\dfrac{2\pi}{T}r\), and so if \(r\) increases (like you see on the graph), so should \(v\)... Unless \(T\) is changing, and messing it all up :P
And that must be the case, here..

Answer 16

^^ Okay :DD Awesome i actually got it too o.O
Thx <3

Answer 17

You're welcome! As for the graph, can anybody here say if it is exponential?
Pinkappellady, have you learned about exponential functions? If not, that's not what your teacher wants. :P

You can say that the relationship is not "linear!" Linear would mean that \(v\) and \(r\) are proportional (like, no funny business of \(v\) being directly proportional to \(r^2\)). Linear means you will see a straight line! But, you don't, so it is not linear. I can explain that if you want. Otherwise, too much information.

So, not linear, and there is some inverse proportionality since one increases and the other decreases.

Answer 18

Not linear \(\rightarrow\) "nonlinear"

But I don't know how else to analyze this! Hopefully someone else can help :)

Answer 19

the data from the figure satisfy the equation \[y = 5.6985x ^{-1}\] ,so for that graph the relation between y and r is inversely proportional

Answer 20

Oh, cool! :)

Answer 21

Inverse proportionality looks like that, I guess! :)
\(x^{-1}\) is fancy for \(\dfrac{1}{x}\).

Answer 22

Here is a graph of inverse proportionality!
http://www.wolframalpha.com/input/?i=y%3D1%2Fx

So, it look like that!

Answer 23

looks*

In that sort of relationship,
\(v=\dfrac{(something\ constant)}{r}\)

Answer 24

\(\implies v\times r=(something\ constant)\)
So, if \(vr\) is always the same number, it is inversely proportional. El-naggar figured out that it was equal to 5.696 every time! You can check for yourself.

Thus, they are inversely proportional. That is your answer!