To all those Maths nerds (angular velocity)

[Hopper]

No offense intended :p

I'm doing a Specialist Maths directed investigation at the moment and am having some serious problems with it. Normally I would ask my teacher but he is a lazy lazy man.

Anyway I am having trouble (can't work out how) working our angular velocity. The part of the investigation I am having trouble with is asking:

"If the angular velocity of the motion is the angle (in radians) through which the object turns in one second, what is the angular velocity of this motion?"

The Differential system is such that:
x'=-y and y'=x
Through the investigaiton I discovered that x=Rcos(t+φ) and y=Rsin(t+φ)


Any help would be greatly appreciated

edit changed an error, x is not equal to y

*For the chance anyone has the Hase and Harris Mathematics for Year 12 Specialist Mats text book, it is on page 321 part 6 of investigation 3*

[sawtell]

two .

[S.]

Are you 100% sure you've typed all that stuff out right? ie x = y = Rcos(blah)?

Edit: that makes more sense. You sure all the other stuff is right, before you get into it?

[Joel O]

"If the angular velocity of the motion is the angle (in radians) through which the object turns in one second, what is the angular velocity of this motion?"

The Differential system is such that:
x'=-y and y'=x
Through the investigaiton I discovered that x=Rcos(t+φ) and y=Rcos(t+φ)
does that not mean x=y? i'm guessing you've noticed this already, i'll keep it in the back of my mind and let you know what i come up with (not giving any answers though).

[Hopper]

Sorry mate, i just picked that up then, it is meant to read
x=Rcos(t+φ) and y=Rsin(t+φ)

[rhyno]

I can hear Steve from here: "Ooh, a maths question!"

[Hopper]

Is the answr 1 radian per second?
as angle at t=0 is 0+t and at t=1 it is 1+t

therefore using final angle - initial angle

radians= (1+t)-(0+t) = 1

I feel that my reasoning is incorrect as it does not take into acount the circular funcitons sinand cosine?

[S.]

Is the answr 1 radian per second?
as angle at t=0 is 0+t and at t=1 it is 1+t

therefore using final angle - initial angle

radians= (1+t)-(0+t) = 1

I feel that my reasoning is incorrect as it does not take into acount the circular funcitons sinand cosine?

You may well be right there actually, but I'd bet that you wouldn't get full marks for that reasoning all the same. Unless you had to actually calculate the equations for X and Y respectively, before this? In which case, that sounds right to me.

[Binaural]

OK, you have the equations for X and Y given to you. What they mean is that given a constant angular velocity, a radius and a time, these equations will tell you the (x,y) co-ordinates of a point. What I think you need to do now is differentiate x=Rcos(t+φ) and y=Rsin(t+φ) with respect to t. R and φ should be treated as constants.

At a guess (been a while), dx/dt = - sin(t+φ) and dy/dt = Rcos(t+φ). Now, since x'=-y and y'=x, you now have two equations you can solve simultaneously to find φ.

so, using the relationship y' = x to find φ (angular velocity):

Rcos(t+φ) = Rcos(t+φ) = 1. Therefore Hopper is right, but there's the formal reasoning for you ;) Hope that helps!

[FuTAnT]

I don't think 1 is the answer. You've proven what was already given, ie x'=y and y'=x by the above equation. So far I've got that ...

Ang Vel = sqrt[r^2*cos(2(t+φ))]

x and y are the position, so x' and y' are the velocities. Thus the actual angular velocity is sqrt(x'^2 + y'^2). When you simplify that you get the above eqn. You can probably simplify it a few different ways, depending on what kind of answer they are looking for.

Does it give you an answer that you're looking to get to?

[FuTAnT]

Do we know what values are assigned to R, t and the angle etc?

Oh yeah, good page for trig identities too ...

http://aleph0.clarku.edu/~djoyce/java/trig/identities.html

[Binaural]

I don't think 1 is the answer. You've proven what was already given, ie x'=y and y'=x by the above equation. So far I've got that ...

Ang Vel = sqrt[r^2*cos(2(t+φ))]

x and y are the position, so x' and y' are the velocities. Thus the actual angular velocity is sqrt(x'^2 + y'^2). When you simplify that you get the above eqn. You can probably simplify it a few different ways, depending on what kind of answer they are looking for.

Does it give you an answer that you're looking to get to?

I think the question is actually asking the reader to prove the obvious. "If the angular velocity of the motion is the angle (in radians) through which the object turns in one second, what is the angular velocity of this motion?" Admittedly the English of the question is so poor it could mean a lot of things, but I am still reasonably confident I am right.

Also, your equation you derived above is φ = sqrt(x'^2 + y'^2). This can be re-arranged to be φ^2 = y'^2 + x'^2 --> y' = sqrt(φ^2 - x'^2). This breaks the relationship y'=x, which means your answer can't be right ;)

Hahaha, love of bit of math with a beer in hand on a monday night!

[Cave Dweller]

Im not sure i understand what your question is :confused:

Are you trying to relate an angular velocity to some xy co-ordinates at a certain radius R?? Or did you find the equations and don't know what they mean??

If the point xy at radius R moves at a constant displacement, velocity or acceleration it will simply follow a simple harmonic wave, or in other words x and y will follow either a sin or cos wave (depending on where you start) and they will be 90 degrees (or pi/2) out of phase with each other (ie one will be sin and the other will be cos).

The angular velocity (in radians) will end up being the natural frequency of the wave. Divide by 2*pi to get it in Hz, invert to get it into seconds.

If that is not what your after and i missed it completely then give us more detail.

[Hopper]

Here is the full question.

I the first 5 parts fully worked out. From what is given, we can see that the angular velocity will be constant for this set of equaitons, speed will increase as further displacement i needed to complete one radian.

Oh and guys thanks so much, you are a huge help guys

[Binaural]

Here is the full question.

I the first 5 parts fully worked out. From what is given, we can see that the angular velocity will be constant for this set of equaitons, speed will increase as further displacement i needed to complete one radian.

Oh and guys thanks so much, you are a huge help guys

OK, looks like I solved question 4 :o I think the answer to question 6 is 2pi radians/s (one rotation per second).

[Cave Dweller]

Okay, ive had a few drinks so go easy on me :p

1. See my answer above, motion will be simple harmonic
2. R will increase amplitude
3. will introduce a time lag
4. I think its R*cos*theta and R*sin*theta
5. blah, do it yourself
6. That is a really badly worded question!! Angular velocity = angle / time? I it moves 1 radian in 1 second angular velocity will be 1rad/s, if it moves 1 full rev it will be 2pi rads like binaural says

[FuTAnT]

Ah, I'm just wondering if φ is actually the angular velocity or not, normally you would just write w (omega).

There's got to be a simple explanation for all of this.

[FuTAnT]

Aha, see, phi is NOT the angular motion. My original equation stands! Cha ching!

[Binaural]

Aha, see, phi is NOT the angular motion. My original equation stands! Cha ching!

God, that was funny! Check my new sig ;)

[FuTAnT]

God, that was funny! Check my new sig ;)

Ha, I'm just chasing myself in circles. Never mind.

[FuTAnT]

Excuse me while I get angular velocity and tangential velocity totally confused.

Anyhoo, w=v/R, v=sqrt(x'^2+y'^2)

Then w=sqrt(cos(2*(t+phi)))

I've got it in a word doc if I can find somewhere to dump it ...

Here we go ... http://members.optusnet.com.au/teamkwaka/trig.doc

So, 2pi is not the answer, unless you decide to assume that the body rotates once in one second, but then that is merely giving you the answer in the question. It is a function of t and phi, both of which need to be selected. The output will give you the angular velocity at any time t around it's motion (non circular).

EDIT: Fixed the eqn. t = time in seconds, phi = a shift from the zero start point (if applicable) and should be in radians, or pi if using deg mode.

[Cave Dweller]

phi = a shift from the zero start point

I do believe i have already said that :p

[FuTAnT]

I do believe i have already said that :p

Yerp, you did, but I didn't read it. Anyhoo ... I'm happy with my solution. And no, the answer isn't 2pi!

[Hopper]

The answer is one!!!!!!!

Reason why:

As the function is a circular function the period must be 2pi seconds.
If the function does one full revolution, it moves through 2pi radians.
Therefore the angualr velocity must be:
amount of radians/time
= 2pi radians/2pi seconds
=1 radian per second :D

[Binaural]

The answer is one!!!!!!!

Reason why:

As the function is a circular function the period must be 2pi seconds.
If the function does one full revolution, it moves through 2pi radians.
Therefore the angualr velocity must be:
amount of radians/time
= 2pi radians/2pi seconds
=1 radian per second :D

That can't be right. velocity = displacement / time
= 2pi radians / 1s
= 2pi radians/s


Where did you get your answer from?

[FuTAnT]

The answer is one!!!!!!!

Reason why:

As the function is a circular function the period must be 2pi seconds.
If the function does one full revolution, it moves through 2pi radians.
Therefore the angualr velocity must be:
amount of radians/time
= 2pi radians/2pi seconds
=1 radian per second :D

Bah, bullshit. The answer isn't 1, it's what I said it was above.

[Ham]

errrm arrrgh its ummmm 42