Help Needed: Deriving Formula for Computer Undergrad's Final Project

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A computer undergraduate is seeking assistance with deriving a formula essential for their final project, which is currently stalled. The mentor has requested the derivation, but the student has struggled for five days without success. Community members are encouraged to provide help, but they request to see the student's previous attempts and identify the most relevant equations. The discussion emphasizes the need for the student to share their work to receive effective guidance. The urgency of the situation highlights the importance of collaboration in academic challenges.
alya
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Homework Statement
Need to derive the following formula trying from the last 5 days
Relevant Equations
https://www.hindawi.com/journals/ijo/2017/4134205/
3d diagram.jpg


formula.jpg


My mentor wants the derivation of this formula.
Me a computer undergrad, unable to figure it out, and my final project are on a halt due to this, any help from the community is greatly appreciated!
 
Last edited:
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alya said:
Homework Statement:: Need to derive the following formula trying from the last 5 days
Relevant Equations:: https://www.hindawi.com/journals/ijo/2017/4134205/

View attachment 282405

View attachment 282404

My mentor wants the derivation of this formula.
Me a computer undergrad, unable to figure it out, and my final project are on a halt due to this, any help from the community is greatly appreciated!
Welcome to PhysicsForums.

Please show us what you have tried so far. We have to see your work before we can offer tutorial help. What do you believe are the most useful Relevant Equations for this problem? (the link you posted is to the full paper about this calculation, not just the several most Relevant Equations)...
 
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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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