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 General Physics General Physics Help Forum Jun 3rd 2017, 07:31 AM #1 Junior Member   Join Date: Dec 2009 Posts: 17 Sun's rotation angular velocity. Hi: I made a formula to find the angular velocity with which the shade rotates about one's position during one day, assuming the Sun rotates with constant angular velocity $\omega$ in a plane at an angle $\alpha$ with respect to the plane of the horizon. I got $\Omega = (1+tan^2 \omega t) \omega cos \alpha / (1+cos^2 \alpha tan^2 \omega t)$ So, $\Omega (0)= \omega cos \alpha$ which makes sense. However $\Omega(90 degrees/ \omega)= \omega /cos \alpha$, which does not. What am I doing wrong? If you need it I can provide the detail of the derivation.   Jun 3rd 2017, 10:07 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 Your formula doesn't work at angle of 90 degrees, because the numerator is infinity and the denominator includes a term that is zero times infinity. I think you're going to have to show us your derivation - perhaps somewhere in it you divided the denominator by a term that is not allowed to be zero.   Jun 3rd 2017, 04:40 PM #3 Junior Member   Join Date: Dec 2009 Posts: 17 Thanks. Let $\Theta$ be the x-y-z coordenates in the plane of the horizon, with the origin $O$ at my position, and z axis pointing to the zenith. $\Pi$ the x'-y'-z' coordinates on the plane of the Sun's "orbit" on the celestial sphere (the ecliptic I think it is called), with the origin at $O$ and intersecting $\Theta$ along the x axis, so angle between $\Theta$ and $\Pi$ is $\alpha$. Assuming the "Sun" rotates in $\Pi$ with angular velocity $\omega$, the position in $\Pi$ reference frame is $s(t)=(cos \omega t, sin \omega t, 0)$. The position in the $Theta$ reference frame is therefor $r(t)=(cos \omega t, cos \alpha sin \omega t, 0)$. Now I just differentiate $r$ to get $\Omega$. Last edited by ENRIQUESTEFANINI; Jun 3rd 2017 at 04:45 PM.   Jun 4th 2017, 06:00 AM #4 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 I see a couple of issues with your approach: 1. The z'-component is not zero - it should be $\displaystyle \sin\alpha \sin\omega t$. 2. The magnitude of r is constant, and should be equal to 1. By making the above correction to the z-component, you will see that the magnitude of r = sqrt(x'^2 + y'^2 + z'^2) does indeed equal 1. 3. I don't understand what you mean about differentiating r. As already noted r is constant, but even if it wasn't, differentiating a length with respect to time yields linear velocity, not angular velocity. By the way, the rate of rotation of the r vector is simply $\displaystyle \omega$.   Jun 5th 2017, 06:10 PM #5 Junior Member   Join Date: Dec 2009 Posts: 17 You mean the shade I cast when I am standing on the street floor rotates about me with constant angular velocity? That's what I want to know! The shade makes an angle $\Omega$ with respect to say, the north-south direction. Let's call this angle $\Omega$. I think its derivative with respect to time, $\lambda(t)$ will be constant but will not be equal to $\omega$. Am I wrong? Furthermore, it will depend on $\alpha$. As you said, there is the term $sin \alpha sin \omega t$ lacking in $r$. My mistake. Last edited by ENRIQUESTEFANINI; Jun 5th 2017 at 06:15 PM.   Jun 6th 2017, 01:08 PM #6 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 Thinking about this some more - I was mistaken, I see now that the angle of your shadow is not the same as ⁡ωt. As an example: if alpha = 90 degrees then the rotation of your shadow is 0 from ⁡ωt = 0 to ⁡ωt = 90, then it switches 180 degrees instantaneously and then the rate becomes zero again. I think it best to consider using spherical coordinates for the position of the sun. To convert from Cartesian to spherical coordinates you use the relationships: $\displaystyle r = \sqrt {x^2 + y^2 + z^2}\\ \theta = \sin^{-1} (\frac z r )\\ \phi = \tan^{-1} (\frac y x )$ The rate of your shadow's rotation on the ground is then simply $\displaystyle \frac {d \phi}{dt}$. After some messy math I get the result $\displaystyle \frac {d\phi}{dt} = \frac { \omega \cos \alpha}{\cos^2 \omega t + \cos^2 \alpha \sin^2 \omega t}$ Because we're dealing with arctan functions, the results are valid only for 0<= ωt <90. And also I should pointy out that your original transformations are valid only on the date of the vernal and autumnal equinoxes (March 21 and September 22), because those are the only dates when the sun rises and sets due east and west. Last edited by ChipB; Jun 6th 2017 at 01:11 PM.  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