Converting between cartesian and polar coordinates

[henryc09]

Homework Statement

Particle is moving with velocity v= ui along the line y=2. What is its v in polar coordinates

Homework Equations

The Attempt at a Solution

I think I'm being really stupid here but not entirely sure where to start. If you integrate to find position you have it as = ut + c i + 2j and then in polar coordinates is this

r=$$\sqrt{}(ut+c)^2 + 4$$r^? But then if you were to differentiate that the velocity would depend on the initial position which can't be right. I'm obviously doing something wrong and haven't got my head round this topic yet, any help would be appreciated.

 

[ehild]

$$r=\sqrt{(ut+c)^2 + 4}$$

The polar angle is:

$$\phi=\arctan(\frac{2}{ut+c})$$

The speed in polar coordinates:

$$v=\sqrt{(dr/dt)^2+( r d\phi /dt)^2 }$$

ehild

 

[kerimek]

I would suggest starting by reviewing the definitions and equations for converting between cartesian and polar coordinates. In cartesian coordinates, the particle's position is described by its x and y coordinates, while in polar coordinates, it is described by its distance from the origin (r) and the angle it makes with the positive x-axis (theta). The equations for converting between the two are:

x = r cos(theta)
y = r sin(theta)

To find the particle's v in polar coordinates, we need to find the velocity components in the r and theta directions. We can do this by differentiating the equations for x and y with respect to time:

v_x = dx/dt = r cos(theta) * dr/dt - r sin(theta) * d(theta)/dt
v_y = dy/dt = r sin(theta) * dr/dt + r cos(theta) * d(theta)/dt

Since the particle is moving along the line y=2, we know that its y velocity (v_y) is constant and equal to 2. Therefore, we can solve for the r and theta components of velocity:

v_x = r cos(theta) * dr/dt - 2
v_y = r sin(theta) * dr/dt + 2

We can then use the Pythagorean theorem to find the magnitude of the velocity (v) in polar coordinates:

v = sqrt(v_x^2 + v_y^2) = sqrt((r cos(theta) * dr/dt - 2)^2 + (r sin(theta) * dr/dt + 2)^2)

And finally, we can use trigonometric identities to find the angle (phi) that the velocity makes with the positive r-axis:

tan(phi) = v_y/v_x = (r sin(theta) * dr/dt + 2)/(r cos(theta) * dr/dt - 2)

Therefore, the particle's velocity in polar coordinates is given by (v, phi). I hope this helps clarify the process of converting between cartesian and polar coordinates and finding the velocity in polar coordinates.