- #1
pedro_crusader
- 1
- 0
- Homework Statement
- I need to understand where the very first equation comes from?
- Relevant Equations
- torque?
Torques acting on a cylinder, with friction
- Thread starter pedro_crusader
- Start date
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- Tags
- Torque
Click For Summary
The discussion focuses on a problem involving a rolling cylinder that appears to be dragging a rod with a plate experiencing friction. Participants express the need for a detailed description of the problem rather than assumptions. There is confusion regarding the meaning of labeled arrows in the diagram, particularly the curvy arrows labeled ##\beta## and ##\omega##. A torque equation is mentioned, but its derivation is unclear without additional context. Clarity on these points is essential for a proper understanding of the mechanics involved.
- #2
- 15,857
- 8,996
Please describe the problem in detail not just part of the proffered solution. What is going on here? It looks like you have a rolling cylinder that is dragging a rod at the end of which is a plate with friction. However, this is only a guess and we do not like guessing if it can be avoided.
Furthermore, what do all these labeled arrows represent, especially the curvy arrows labeled ##\beta## and ##\omega##? Again, we do not like guessing.
That said, yes, the first equation looks like a torque equation. Where it comes from depends on details that we do not have.
Furthermore, what do all these labeled arrows represent, especially the curvy arrows labeled ##\beta## and ##\omega##? Again, we do not like guessing.
That said, yes, the first equation looks like a torque equation. Where it comes from depends on details that we do not have.
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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