Kinematics in Polar Coordinates

[theloathedone]

Homework Statement

A particle starts at (d, 0) in polar coordinates and has a velocity of
\vec{v}=(u \sin{\theta} - v)\hat{r} + u \cos{\theta} \hat{\theta}
where v > u

Find the position vector of the particle as a function of time.

Homework Equations

\vec{v}=\frac{d\vec{r}}{dt}

The Attempt at a Solution

I'm not sure how to integrate the expression for v to get the position of a particle. Since the unit vectors are changing, I can't just integrate the components individually like in Cartesian coordinates right?

I also read that \dot{\textbf{r}}=\dot{r}\hat{\textbf{r}}+ r\dot{\theta}\hat{\textbf{\theta}}

But I'm not sure if this helps me.

Thanks in advance for the help :)

P.S. How do I bold the letters? I used \textbf but it doesn't seem to have an effect.

 

[ehild]

The vector equations are equivalent to a system of scalar differential equations:

dr/dt=u sin(θ)-v; r dθ/dt = u cos (θ).

Are u and v constant parameters?

ehild

 

[theloathedone]

Yup u and v are constant.

Is this set of DE solvable?

Edit: After looking at another problem from this forum, I think it would be possible to express it in terms of dr/dθ.

\frac{dr}{d\theta} = \frac{dr}{dt} \frac{1}{\frac{d\theta}{dt}} = ur \tan{\theta} - \frac{vr}{u\cos{\theta}}

And this gives a closed form solution for r(θ). But will I be able to get the position as a function of time for that?

 

[ehild]

\frac{dr}{d\theta} = \frac{dr}{dt} \frac{1}{\frac{d\theta}{dt}} = ur \tan{\theta} - \frac{vr}{u\cos{\theta}}

And this gives a closed form solution for r(θ). But will I be able to get the position as a function of time for that?

Correctly, it is dr/dθ =r tan(θ)-(r v)/(u cos (θ)).
Yes, this has a solution in closed form (I am still struggling with it) and in principle, you should get both r and θ as functions of time from it: if r(θ) is known, dθ/dt=u cos(θ)/r(θ), which is a separable differential equation, and once θ(t) is known you get r(t) by integrating u sin(θ(t))-v.
But it is just ugly...

ehild

 

[theloathedone]

Yeah it is ugly, but I'm allowed to use numerical integration.
Oh yeah I made a mistake in the signs, all of them should have been negative in the original expression for the velocity.

So after integrating, I got r(θ) = K cos(θ)|sec(θ)+tan(θ)|^(v/u)

The particle starts at (d, 0), and I just let d = 100 m, v = 7 m/s, u = 5 m/s (random parameters from the original problem statement). It starts off headed southwest, taking a long arc that looks vaguely semielliptical before reaching the origin.

I want to find the time taken to reach the origin. Do I integrate dθ from 2Pi to 3Pi/2? Since the particle's initial position is at (d, 2Pi) and just before touching the origin, since it is coming northward from the bottom, it will be at (almost 0, 3Pi/2). Is this reasoning correct?

Thanks for the help so far :)

 

[theloathedone]

Ok I tried using Wolfram Alpha to do the numerical integration.

\frac{d\theta}{dt}=\frac{u \cos{\theta}}{r(\theta)} = \frac{u}{K}|\sec{\theta}+tan{\theta}|^{-\frac{v}{u}}

So can I just separate the variables and get dθ|sec(θ)+tan(θ)^(v/u) = u/K dt?

When I used Wolfram Alpha to perform the numerical integration, I get t = 5.58. But this is clearly unreasonable. The maximum speed the particle takes is when sin(theta) = 0 (because it is negative within 3pi/2 to 2pi) and is only √(u^2 + v^2) = 8.60 m/s. To travel the 100m to the origin in a straight line path at that speed would take 11 seconds, so the answer is clearly wrong.

Is there a fundamental flaw in my reasoning somewhere?

Thanks :)

 

[Lostinthought]

convert to cartesian usin
r=cos8i+sin8j
8=sin8i-cos8j
integrate seperatly to get x and y coordinates

 

[ehild]

Is there a fundamental flaw in my reasoning somewhere?

I got an other solution for r(θ) than yours: integral(tan(θ)dθ) is - ln(cos(θ)), so cos(θ) is in the denominator. Check it!

ehild

 

[ehild]

convert to cartesian usin
r=cos8i+sin8j
8=sin8i-cos8j
integrate seperatly to get x and y coordinates

What do you mean with those 8-s?

I tried, it is not easier in x,y coordinates.


ehild

 

[Lostinthought]

i ment <br /> \theta<br />

 

[theloathedone]

Thanks ehild for pointing out my mistake :)

I seem to be making a lot of careless mistakes on this problem.

So after correcting that cos in the denominator, I managed to get answers that sound reasonable, like 30 seconds. I think it should be right now.

Thanks for all the help! Really appreciate your taking time to do the integral etc. and show me where I went wrong :)

 

[ehild]

The pleasure was mine Really, it was a challenging problem.

ehild