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Lets say i am using a counterweight and a weight on the other end of an arm (that i want to launch like a trebuchet)- im not sure how to find the formula between the range of motion and the weight (and angle of launch): this is what i tried-

$$Ek=Ep \text{(energy converted from potential to kinetic of weight)}$$ $$\frac{mv^2}{2}=mgh, v= \sqrt{2gh}$$

Then i used $Range= v^2\frac{\sin(2\theta)}{g}$ to find range but i keep getting really weird answers- where am i going wrong with this please?

NMech
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Johnny
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    You will have the moment of inertia (and frictions) countering the torques causing rotation. The rotational speed achieved (even through multiple pivots) and the tangential point of release will define the range. – Jim Clark Oct 26 '21 at 16:17
  • Since the launched object **and** the counter weight are both moving with non zero velocity at the instant the object leaves the mechanism, only a **fraction** of the initial potential energy of the counter weight is transferred to the launched object as kinetic energy. Moreover, depending on the size and shape mechanism, the potential energy gained by the launched object just before leaving the mechanism also needs to be accounted. – AJN Oct 27 '21 at 02:03
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    A neat diagram clearly showing **all** the symbols used in the expressions above will be helpful for analysing the situation. Possibly two neat diagrams; one showing initial configuration and one showing the configuration just before the launched object leaves the mechanism. – AJN Oct 27 '21 at 02:05
  • @AJN your conclusion is right, but many designs like this determine the launch angle with a stop on the counterweight, so rather than the counterweight carrying some energy it dumps the energy into pounding the stop. – Chris H Oct 29 '21 at 14:57

1 Answers1

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There is a detailed explanation on "Real-world-physics-problems.com".

Here is the Link, trebuchet

kamran
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