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- Thread starter kamil.borkowski
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DaveC426913

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berkeman

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Welcome to the PF.

Is this for launching clay "pigeons" like in trap shooting? Is it a flat disk or a round ball that you are launching. If it's a flat disk like a clay pigeon, you will want it to roll a ways down the launch blade as the blade is going through its motion. That will give it the spin it needs to fly stably.

Is this a spring-powered launcher? If so, you may be able to do a pretty simple calculation equating the energy stored in the spring when cocked with the final energy of the moving arm and the moving target (linear velocity and energy stored in its rotation)...

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_{0}? I see it in the diagram, but what is it doing there? Is there more than one moving component?

r

Welcome to the PF.

Is this for launching clay "pigeons" like in trap shooting? Is it a flat disk or a round ball that you are launching. If it's a flat disk like a clay pigeon, you will want it to roll a ways down the launch blade as the blade is going through its motion. That will give it the spin it needs to fly stably.

Is this a spring-powered launcher? If so, you may be able to do a pretty simple calculation equating the energy stored in the spring when cocked with the final energy of the moving arm and the moving target (linear velocity and energy stored in its rotation)...

Yes, it is a clay launcher/thrower. It is spring-powered. I was thinking about the energy calculations, but I thought that it would leave the aspect of WHEN the target leaves the arm. I want it to leave, when the arm travels about 160°-170°. In energy equation I need also ω of the arm and that's the problem. I can't find proper ω of the arm.

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DaveC426913

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Ohhhh I see. So it follows a spiral path.r_{0}is a path, that disc has to travel, from it's initial position (V_{0}=0), to the moment when it leaves the arm (V=27)

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Ohhhh I see. So it follows a spiral path.

Yes, exactly! I'm using that equation: E=1/2*(m*V^2 + Iω^2 + Iω^2). I know V and ω of disc, but I don't know how to calculate ω of the arm.

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berkeman

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Yes, I know, but I have to calculate it somehow (ω of the arm), because I need it to equation of energy.

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berkeman

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Well, maybe not full solution. The thing is I need to choose proper spring. I will do it from energy equation, just like You sugested. To do so, I need angular and linear velocity of clay target (which I have) and angular velocity of the arm. And that's the problem, how to calculate ω of the arm. I can't find connection between ω of arm and velocities of clay target. Not to mention, that the friction channel is complicating the case even more, cuz I don't have an idea how to take it to account neither.

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DaveC426913

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Thanks Dave, I was thinking in other system, and I see the sketch shows the two position.

Kamil, I have not found an easy solution: two second order nonlinear differential equations, which, of course, there would be possible to solve numericaly .

x is the distance target-center of the bar

v es the speed of the target which is the vectorial sum of the two ortogonal vectors: \dot{x} and \dot{α}.x

k is the spring constant

The two differential equations are:

(1). \ddot{α}.(I+m.x)=k.α

(2). \ddot{x}=\dotα^{2}.x

Sorry i don't know how to write the equations properly. I'll try to solve it.

Kamil, I have not found an easy solution: two second order nonlinear differential equations, which, of course, there would be possible to solve numericaly .

x is the distance target-center of the bar

v es the speed of the target which is the vectorial sum of the two ortogonal vectors: \dot{x} and \dot{α}.x

k is the spring constant

The two differential equations are:

(1). \ddot{α}.(I+m.x)=k.α

(2). \ddot{x}=\dotα

Sorry i don't know how to write the equations properly. I'll try to solve it.

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