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3Dimensional Projectile Motion 
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#1
Apr705, 02:30 PM

P: 666

I know in 2dimensions, the x coordinate is represented by
[tex]x=v_{0}\cos{(\theta)}t,[/tex] and the y coodinate is represented by [tex]y=\frac{1}{2}gt^2+v_{0}\sin{(\theta)}t+h.[/tex] How would you calculuate the z coordinate if it was rotating around the y axis? For example; a sprinkler. Thanks for your help. 


#2
Apr705, 03:05 PM

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Zz. 


#3
Apr705, 03:33 PM

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Who's rotating??
Daniel. 


#4
Apr705, 04:23 PM

P: 666

3Dimensional Projectile Motion
What would you use? Even if there was a better way, I am interested in how it would be defined in rec. coordinates.
Thanks again. 


#5
Apr705, 04:52 PM

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Are u thinking of a spinning (finite size) projectile wondering through the (viscous,moving,nonisothermal,nonisobaric) atmosphere,in the nonconstant nonhomogenous gravitational field created by a rotating Earth??
Daniel. 


#6
Apr705, 07:07 PM

P: 666

I'm thinking of no outside forces besides gravity.



#7
Apr705, 07:10 PM

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Anyway,a spinning rigid body is typically discribed by 6 coordinates:the 3 cartesian for the CM (parametrize the body's translation) and the 3 Euler angles (parametrize the body's rotations).
Daniel. 


#8
Apr705, 07:11 PM

P: 666

Ok well, I haven't had a physics class before, so what I know is strictly what I have read out of a book (which isn't much). What type of physics class would I learn these types of things in?



#9
Apr705, 07:15 PM

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A college course in classical mechanics in Newtonian formulation.
Daniel. 


#10
Oct1010, 11:19 AM

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#11
Oct1010, 11:23 AM

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I'd love to hear your answer on this as I too was wondering why Newtons EK Kinetic Energy does not include rotating or orbital objects. Keep me posted? thanks MJL 


#12
Oct1010, 03:50 PM

P: 168

How about K = (1/2)Iw^2. Where K is the angular kinetic energy, I is the moment of inertia of the rotating object, and w, it should be omega, is the angular velocity.
If you are trying to analyze the angular kinetic energy of water that leaves a sprinkler I can help you out there. It is zero. Once it leaves the jets it goes in a straight line. If you want to analyze what happens to the water once it leaves the jets it is best done by getting its velocity at the moment it leaves. Then you reduce it to a two dimensional analysis using y = (1/2)at^2 + vtsin(theta) + c, and x = vcos(theta) + d. Where c and d are the initial y and x values and v is the initial velocity. The energy analysis of the rotating sprinkler is a bit more complex. It all depends on the shape of it and its mass distribution. But from your question I believe that you wanted to analyze the water once it left the sprinkler, I may be wrong. 


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