Spiral motion and time

[Jiralhanae125]

This is a little complicated, so i'll set up the premise here:
I have a logarithmic spiral with polar equation r = k e^-bθ where k and b are constants, b > 0.

From simple trig, and conversion to cartesian form, I found the parametric equations:
x = r cosθ
y = r sinθ

substituting for r above gives:
x = k e^-bθcosθ
y = k e^-bθsinθ

Now I am stuck in introducing the element of time into the equation. I know that the angular velocity of an object is 2π/twhere tis the time taken for the object to complete a revolution. This is where the problem occurs: I know that logically, in spiral motion as the value of r decreases, t decreases, which increases the angular velocity. Now how on earth do I express the tin terms of r without making something recursive? I could express t in terms of θ but then [FONT=arial, sans-serif]θ would be dependent on the angular velocity and time which would in turn be dependent on the t and so the cycle continues... any ideas?[/FONT]

[FONT=arial, sans-serif]Also any idea on how to find the distance traveled by an object is spiral motion over different time or position parameters?[/FONT]

 

[Ishuda]

Whether r decreases with t (time) or increases with t depends on whether the spiral is an inward or outward spiral. The angular velocity will depend on the relationship between t and \(\displaystyle \theta\). If it is a straight linear relationship then, since angular velocity is \(\displaystyle d\theta/dt\), the angular velocity is constant. For constant angular velocity, the angular velocity is sometimes expressed as \(\displaystyle \omega = 2 \pi f\) where f is the frequency of rotation, i.e. once per second would be a frequency of 1 hertz. The tangential velocity, r \(\displaystyle d\theta/dt\), does change with r and thus time.

Path length is the integral
\(\displaystyle \int ds = \int [dx^2 + dy^2]^{1/2} = \int [1 + (dy/dx)^2]^{1/2} dx\)
Since the integral is along the path, it may need to be broken up if you do it in Cartesian co-ordinates, e.g. think of a circle.

 

[Jiralhanae125]

Whether r decreases with t (time) or increases with t depends on whether the spiral is an inward or outward spiral. The angular velocity will depend on the relationship between t and \(\displaystyle \theta\). If it is a straight linear relationship then, since angular velocity is \(\displaystyle d\theta/dt\), the angular velocity is constant. For constant angular velocity, the angular velocity is sometimes expressed as \(\displaystyle \omega = 2 \pi f\) where f is the frequency of rotation, i.e. once per second would be a frequency of 1 hertz. The tangential velocity, r \(\displaystyle d\theta/dt\), does change with r and thus time.

Path length is the integral
\(\displaystyle \int ds = \int [dx^2 + dy^2]^{1/2} = \int [1 + (dy/dx)^2]^{1/2} dx\)
Since the integral is along the path, it may need to be broken up if you do it in Cartesian co-ordinates, e.g. think of a circle.

Thanks, but what happens if angular velocity is not constant, i.e. it increases with time?
How can I express a changing dθ/dt that is not dependent on radius (because radius is dependent on θ)

 

[Ishuda]

Thanks, but what happens if angular velocity is not constant, i.e. it increases with time?
How can I express a changing dθ/dt that is not dependent on radius (because radius is dependent on θ)

You need to know the relationship between \(\displaystyle \theta\) and t before you can compute angular velocity. As the problem is given, there is no such relationship stated. The angular velocity of an object is NOT (in general) 2π/t where t is the time taken for the object to complete a revolution. That value, 2π/t, is the average angular velocity over one revolution and the average frequency is then 1/t hertz (assuming t was in seconds).

Suppose the angular velocity decreased exponentially as t increased, i.e. we had something like
\(\displaystyle \theta = \theta_0 + k\space (1 - e^{-\dfrac{\omega_0}{k}t}\))
so that
\(\displaystyle \dfrac{d\theta}{dt}=\space \omega_0\space e^{-\dfrac{\omega_0}{k}t}\)
You could still have you logarythmic spiral as you have described it. It would be connected with time through the above equation.

EDIT: Correct stupid misteak

 

[Jiralhanae125]

You need to know the relationship between \(\displaystyle \theta\) and t before you can compute angular velocity. As the problem is given, there is no such relationship stated. The angular velocity of an object is NOT (in general) 2π/t where t is the time taken for the object to complete a revolution. That value, 2π/t, is the average angular velocity over one revolution and the average frequency is then 1/t hertz (assuming t was in seconds).

Suppose the angular velocity decreased exponentially as t increased, i.e. we had something like
\(\displaystyle \theta = \theta_0 + k\space (1 - e^{-\dfrac{\omega_0}{k}t}\))
so that
\(\displaystyle \dfrac{d\theta}{dt}=\space \omega_0\space e^{-\dfrac{\omega_0}{k}t}\)
You could still have you logarythmic spiral as you have described it. It would be connected with time through the above equation.

Ok this is making more sense to me. I have two final questions. Firstly, in the equation above, how would one calculate ω_{0 ,} or is that just an assumed constant? And would this method work: suppose I decide to use Kepler's 3rd (t²/r³= constant) Then I can find dr/dt. Using the parametric equations in my original post I can find dθ/dr
. So then I divide (dr/dt) / (dr/dθ) to get dθ/dt. Then I differentiate again to get the change in dθ/dt. Or would I end up getting dθ/dt as 2pi/t always?Thanks for your help so far by the way, my math teacher was not much help to me on this!

 

[Ishuda]

Ok this is making more sense to me. I have two final questions. Firstly, in the equation above, how would one calculate ω_{0 ,} or is that just an assumed constant? And would this method work: suppose I decide to use Kepler's 3rd (t²/r³= constant) Then I can find dr/dt. Using the parametric equations in my original post I can find dθ/dr
. So then I divide (dr/dt) / (dr/dθ) to get dθ/dt. Then I differentiate again to get the change in dθ/dt. Or would I end up getting dθ/dt as 2pi/t always?Thanks for your help so far by the way, my math teacher was not much help to me on this!

If you decided to go that way with your original equation
r = k e^-bθ
there would be no reason to go to that much trouble. That is, let c be a constant (the inverse of your Kepler's 3rd constant) and
r³ = c t²
Then
c t² = [k e^-bθ]³ = k³ e^{-3b\(\displaystyle \theta\)}
or, rearranging
\(\displaystyle \theta\) = \(\displaystyle \alpha\) ln(t) + \(\displaystyle \beta\)
and
\(\displaystyle \dfrac{d\theta}{dt} = \alpha\space t^{-1}\)

If you can isolate r as a function of t, i.e.
r = \(\displaystyle c^{\frac{1}{3}}\space t^{\frac{2}{3}}\)
there is no need to go through that trouble although you certainly could.

About your statement "Or would I end up getting dθ/dt as 2\(\displaystyle \p\)i/t always?" That is only true if \(\displaystyle \theta\) is linear in time. Again, the expression \(\displaystyle \dfrac{2\space \pi}{t_0}\) where t₀ is the time for one revolution is the average angular velocity for that revolution.

 

[Jiralhanae125]

If you decided to go that way with your original equation
r = k e^-bθ
there would be no reason to go to that much trouble. That is, let c be a constant (the inverse of your Kepler's 3rd constant) and
r³ = c t²
Then
c t² = [k e^-bθ]³ = k³ e^{-3b\(\displaystyle \theta\)}
or, rearranging
\(\displaystyle \theta\) = \(\displaystyle \alpha\) ln(t) + \(\displaystyle \beta\)
and
\(\displaystyle \dfrac{d\theta}{dt} = \alpha\space t^{-1}\)

If you can isolate r as a function of t, i.e.
r = \(\displaystyle c^{\frac{1}{3}}\space t^{\frac{2}{3}}\)
there is no need to go through that trouble although you certainly could.

About your statement "Or would I end up getting dθ/dt as 2\(\displaystyle \p\)i/t always?" That is only true if \(\displaystyle \theta\) is linear in time. Again, the expression \(\displaystyle \dfrac{2\space \pi}{t_0}\) where t₀ is the time for one revolution is the average angular velocity for that revolution.

Ah I think I see a flaw in the Kepler's third law. the time t is actually capital T meaning the orbital period, which is not the same as t for the time the object has been travelling. So I'm not sure my calculations were working. I think i'll stick with your equations for varying orbital velocity. My final question is whether the constants k you use in that equation is the same as the constant that I use in my equation for r = k e^-bθ

 

[Ishuda]

...My final question is whether the constants k you use in that equation is the same as the constant that I use in my equation for r = k e^-bθ

Yes, I was using your example for r = k e^-bθ so it was the same k.

Oh, and I noticed I didn't answer your question "... how would one calculate ω₀, or is that just an assumed constant?" It was just an assumed constant.

 

[Jiralhanae125]

Yes, I was using your example for r = k e^-bθ so it was the same k.

Oh, and I noticed I didn't answer your question "... how would one calculate ω₀, or is that just an assumed constant?" It was just an assumed constant.

Why would it be the same constant k for angular velocity as it would for the radius?

 

[Ishuda]

Why would it be the same constant k for angular velocity as it would for the radius?

The equations you had was the original
r = k e^-bθ
and the (mislabeled Kepler's 3rd) added equation
r³ = c t²
Thus
r³ = [k e^-bθ]³ = c t²
so that
ln ([k e^-bθ]³] = 3 ln(k) - 3b \(\displaystyle \theta] = ln(c t²) = ln(c) + 2 ln(t)
or
\(\displaystyle \theta = \frac{3\space ln(k) - ln(c)}{3b} - \frac{2}{3b}\space ln(t) = \alpha t + \beta\)
where
\(\displaystyle \alpha = - \frac{2}{3b}\)
and
\(\displaystyle \beta = \frac{3\space ln(k) - ln(c)}{3b}\)\)