# Help with a curve problem

1. Mar 25, 2004

### FabioTTT

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?

2. Mar 25, 2004

### 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.

3. Mar 25, 2004

### 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.

Last edited: Mar 25, 2004
4. Mar 25, 2004

### 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.

5. Mar 25, 2004

### 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.

6. Mar 25, 2004

### 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.

7. Mar 25, 2004

### FabioTTT

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.

8. Mar 26, 2004

### FabioTTT

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.

Last edited: Mar 26, 2004
9. Mar 27, 2004

### 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).

10. Mar 27, 2004

### 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

Last edited: Mar 27, 2004
11. Mar 27, 2004

### Muzza

12. Mar 27, 2004

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, $$2\pi r^2$$