- #1
physicss
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- Homework Statement
- Hello, I had to calculate ∆r and ∆Theta,phi. Is the answer on the second picture correct?
- Relevant Equations
- ∆r, ∆Theta,phi
Laplace operator in spherical coordinates
- Thread starter physicss
- Start date
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- Tags
- Laplace
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The discussion revolves around the confusion regarding the labeling of images related to the Laplace operator in spherical coordinates. Participants emphasize the importance of clear labeling to avoid misunderstandings. A link to the Wikipedia page on the topic is provided as a resource for clarification. There is also a note about the readability of images, specifically advising against posting them sideways. Clear communication and proper formatting are highlighted as essential for effective discussion.
- #2
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It's not clear which is the first and which is the second picture. You can find all you need here
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
Please don't post pictures sideways. They are hard to read. Thanks.
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
Please don't post pictures sideways. They are hard to read. Thanks.
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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