Coordinate systems and velocitiesby Spanky Deluxe Tags: cartesian, conversion, coordinate systems, cylindrical, velocity 

#1
Jul709, 06:48 PM

P: 19

Can anyone help explain how to convert velocities to different co ordinate systems? I've always been shocking at converting between co ordinate systems as it is.
I know that converting from Cartesian to Cylindrical for positions it is: r = sqrt(x^2 + y^2) theta = atan(y/x) z = z and back its: x = rcos(theta) y = rsin(theta) z = z However, I'm getting confused as to what it would be for velocities. What I've got so far is: vx = vtheta * sin(theta) * sign(y) vy = vtheta * cos(theta) * sign(x) vz = vz Am I doing the right kind of thing or am I overcomplicating things? Any help would be greatly appreciated!! 



#2
Jul709, 06:57 PM

P: 1,909

What is "vtheta'?
x is a function of two variables, r and theta and these are both functions of time, in general. dx/dt=(dr/dt)cos(theta)r sin(theta) d(theta)/dt and similar for y. dr/dt is the radial speed and d(theta)/dt is angular speed. 



#3
Jul709, 07:07 PM

P: 19

vtheta is the rotational velocity




#4
Jul809, 07:46 AM

Mentor
P: 16,485

Coordinate systems and velocities
Hi Spanky,
I think you are overcomplicating things. If you have any vector in Cartesian coordinates then to transform it to Cylindrical coordinates you use r = sqrt(x^2 + y^2) theta = atan(y/x) z = z That part is easy. What is more challenging is determining the velocity vector in Cylindrical coordinates if you have a position in Cylindrical coordinates as a function of time. For that, see the "time derivatives of the vector" section here: http://mathworld.wolfram.com/Cylindr...ordinates.html v = (r', r theta', z') 



#5
Jul809, 02:26 PM

P: 19

I think I'm getting muddled up with different coordinate systems, rotational velocities and the spin dependence of elephants when eating carrot sticks.
Looking at everything again, I basically need to be able to convert between azimuthal and radial velocities and x and y velocities for a body rotating around another body. Its all basically a mess of those diagrams I had to draw at school working out different angles and their dependence on each other, which I sucked donkey doo doo at. I basically need to convert backwards and forwards between azimuthal and radial components of force and velocity to Cartesian coordinates. The origin is at (0,0). So I guess I need something like this for anticlockwise motion: theta = (y/x) vx = (v_azimuthal * sin(theta) * sign(y)) + (v_radial * cos(theta) * sign(x)) vy = (v_azimuthal * cos(theta) * sign(x)) + (v_radial * sin(theta) * sign(y)) vz = vz Then the same should be true for forces I think. For clockwise motion the first sign functions of vx and vy would need to be sign(y) and sign(x) respectively I think. I think those equations are right, if a bit complex. Now though, I need to work out a way to get v_azimuthal and v_radial from vx and vy, i.e. the other way around. 



#6
Jul809, 04:01 PM

Mentor
P: 16,485

If you have any vector in Cartesian coordinates and you want to express it in Cylindrical coordinates then you use:
r = sqrt(x²+y²) theta = atan(y/x) z = z If you have any vector in Cylindrical coordinates and you want to express it in Cartesian coordinates then you use: x = r cos(theta) y = r sin(theta) z = z It makes no difference whatsoever what the units are nor whether the vector represents position, acceleration, velocity, momentum, etc. The only trick comes if you have a position as a function of time in Cylindrical coordinates and you want to find the velocity or acceleration as a function of time in the Cylindrical coordinates without converting to and from Cartesian. Then you use the formulas I linked to above. 



#7
Jul809, 04:04 PM

HW Helper
P: 5,004

Why are there sign functions in your expression?
To find velocities you need only differentiate withe respect to time: [tex]v_x\equiv\frac{dx}{dt}=\frac{d}{dt}(r\cos\theta)=\frac{dr}{dt}\cos\thet ar\sin\theta\frac{d\theta}{dt}=v_r\cos\theta v_{\theta}\sin\theta[/tex] To get [itex]v_y[/itex] and [itex]v_z[/itex] you just do the same thing to your expressions for [itex]y[/itex] and [itex]z[/itex] To get [itex]v_r[/itex] and [itex]v_{\theta}[/itex] in terms of Cartesians you just differentiate the expressions for [itex]r[/itex] and [itex]\theta[/itex] with respect to time: [tex]v_r=\frac{dr}{dt}=\frac{d}{dt}\sqrt{x^2+y^2}[/tex] [tex]v_{\theta}\equiv r\frac{d\theta}{dt}=r\frac{d}{dt} \tan^{1}\left(\frac{y}{x}\right)=\sqrt{x^2+y^2}\frac{d}{dt} \tan^{1}\left(\frac{y}{x}\right)[/tex] OR you solve the 3 equations you got for v_x, v_y and v_z for v_r, v_theta and v_z the same way you would solve any system of 3 equations for 3 unknowns. 



#8
Jul809, 05:41 PM


#9
Jul809, 10:33 PM

Mentor
P: 16,485




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