help with a curve problem

[FabioTTT]

http://seminario.no-ip.com/problem2.gif

I'm confused about the above problem because I don't know why there would be an a,b, or greek letter "w" in there. Anyone?

[matt grime]

They're constants. That's all. Position is still a function of t. w might well work out to be something like angular momentum.

[Chen]

Looks like helical motion... the kind of motion you would if you gave a particle an inital velocity inside a magnetic field, when there's angle between the velocity and the field.

In the XY plane the particle is going in a circle, the radius of which is a, and the angular velocity of the particle in that plane is \omega (omega). The velocity in that plane is:
v_r = \omega a

Along the Z axis, the particle is traveling at a constant speed of b. Since the velocities in the Z axis and in the XY plane are perpendicular, the magnitude of the velocity of the particle is:
V = \sqrt{b^2 + v_r^2} = \sqrt{b^2 + \omega ^2a^2}

As for the acceleration, it only exists in the XY plane (since the motion along the Z axis is at a constant speed). The acceleration in a circular motion is given by:
a_r = \omega ^2a

The angle \theta between the velocity and the acceleartion, is the angle between the velocity and the XY plane.

[verty]

'w' (that fancy w) is angular velocity, and is measured in radians per second. 'w = 2*pi*f', where f is the frequency in Hz.

[FabioTTT]

Originally posted by Chen
Looks like helical motion... the kind of motion you would if you gave a particle an inital velocity inside a magnetic field, when there's angle between the velocity and the field.

In the XY plane the particle is going in a circle, the radius of which is a, and the angular velocity of the particle in that plane is \omega (omega). The velocity in that plane is:
v_r = \omega a

Along the Z axis, the particle is traveling at a constant speed of b. Since the velocities in the Z axis and in the XY plane are perpendicular, the magnitude of the velocity of the particle is:
V = \sqrt{b^2 + v_r^2} = \sqrt{b^2 + \omega ^2a^2}

As for the acceleration, it only exists in the XY plane (since the motion along the Z axis is at a constant speed). The acceleration in a circular motion is given by:
a_r = \omega ^2a

The angle \theta between the velocity and the acceleartion, is the angle between the velocity and the XY plane.

Thanks for the info... it's funny this problem is nothing like any of teh other problems i've encountered in this section of my math text book.

[Chen]

Originally posted by FabioTTT
Thanks for the info... it's funny this problem is nothing like any of teh other problems i've encountered in this section of my math text book.
I didn't notice this is a math problem... it does include some physics but you can also prove it mathematically.

For example, you know that the change of position is velocity right? So if:
x = a\cos (\omega t)
y = a\sin (\omega t)
Derive both to get the velocity in each axis:
x' = v_x = -\omega a\sin (\omega t)
y' = v_y = \omega a\cos (\omega t)
To find the "total" velocity in the XY plane, use pythagoras:
v_{xy}^2 = v_x^2 + v_y^2 = \omega ^2a^2(\sin ^2(\omega t) + \cos ^2(\omega t) = \omega ^2a^2
Therefore:
v_{xy} = \omega a
Just like I said above, but here you actually proved it. [:)]

[FabioTTT]

Originally posted by Chen
I didn't notice this is a math problem... it does include some physics but you can also prove it mathematically.

For example, you know that the change of position is velocity right? So if:
x = a\cos (\omega t)
y = a\sin (\omega t)
Derive both to get the velocity in each axis:
x' = v_x = -\omega a\sin (\omega t)
y' = v_y = \omega a\cos (\omega t)
To find the "total" velocity in the XY plane, use pythagoras:
v_{xy}^2 = v_x^2 + v_y^2 = \omega ^2a^2(\sin ^2(\omega t) + \cos ^2(\omega t) = \omega ^2a^2
Therefore:
v_{xy} = \omega a
Just like I said above, but here you actually proved it. [:)]

thanks. I'm aware of how to obtain the velocity, speed and acceleration vectors, however I just didnt know why he would include a,b,w. doesnt make much sense to me.

[FabioTTT]

I didn't notice this is a math problem... it does include some physics but you can also prove it mathematically.

For example, you know that the change of position is velocity right? So if:
x = a\cos (\omega t)
y = a\sin (\omega t)
Derive both to get the velocity in each axis:
x' = v_x = -\omega a\sin (\omega t)
y' = v_y = \omega a\cos (\omega t)
To find the "total" velocity in the XY plane, use pythagoras:
v_{xy}^2 = v_x^2 + v_y^2 = \omega ^2a^2(\sin ^2(\omega t) + \cos ^2(\omega t) = \omega ^2a^2
Therefore:
v_{xy} = \omega a
Just like I said above, but here you actually proved it. [:)]

question, i understand the way you worked this, however, why would i have to calculate the velocities separately for each axis?

wouldnt the position vector be r(t) = acos(wt)i + asin(wt)j + btk ?

then i would just do dr/dt to find the velocity vector wouldnt i? and dv/dt to find the acceleration?

im only questioning this because if i do it the way i mentioned above, for speed i end up with

sqroot( w^2 a^2 + b^2 ) which doesnt seem right since it would make the rest of the questions quite messy. but i've checked it and double checked it and thats what I come up with.

[Chen]

The speed is indeed \sqrt{\omega ^2a^2 + b^2}, you got it exactly right (this is also what I said in post #3).

[FabioTTT]

can i ask what programs you guys use to write out your mathematical formulas? i realize they are images.. so you must be making them some where.

edit: by the way, thanks chen

[Muzza]

FabioTTT, they're generated "on-the-fly" by LaTeX. See http://www.physicsforums.com/showthread.php?t=8997 for more information (I wonder why that thread isn't stickied in more fora).

[selfAdjoint]

We use LaTex. There are stickies around which tell you how to format. These aren't images but are created inline thus [ tex ] 2\pi r^2 [ / tex] becomes, if you remove the blanks, 2\pi r^2

[FabioTTT]

oh ok, thanks