Is Throwing a Knife with No Spin Actually More Effective?

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The discussion centers on the effectiveness of throwing a knife with no spin compared to a rotating throw. It highlights that a spinning knife experiences more air drag, particularly when rotated horizontally, which could affect its trajectory. Some argue that a no-spin throw may allow for higher velocity and better accuracy, as the knife's end hitting the target may not be predictable with a spinning motion. Participants request more scientific sources to support claims made about knife throwing techniques, emphasizing the need for credible references over anecdotal evidence. Overall, the debate underscores the complexities of knife throwing dynamics and the importance of reliable information.
Aurelius120
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Homework Statement
I have heard that if you spin a knife it hits harder than if allowed to fly tangentially
Relevant Equations
NA
They say that a rotating knife thrown is more dangerous than a knife thrown straight

I find it weird

  • If the knife is rotating, it will experience more air drag than if thrown straight which will also depend on plane of rotation(For some reason, I don't know, it experiences more drag if rotated in horizontal plane than if in the vertical plane)
  • One end of it will have greater velocity than it could have in pure translation but hey! There is no saying which end will hit the target and that end will probably have maximum velocity(as in rolling motion)
So does that actually mean any advantage
 
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It is easier to get a high velocity with a throw that spins the knife from the point instead of one that throws it like a dart.
 
Aurelius120 said:
They say
Can you post a link to where you have read this please? As @haruspex says, there are other considerations...
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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