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YaniHozya
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Rolling pendulum angular acceleration
- Thread starter YaniHozya
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To solve the problem of rolling pendulum angular acceleration, it's essential to demonstrate an attempt at a solution. Participants are encouraged to reference specific methods taught in their module related to the topic. Sharing any preliminary work or calculations is crucial for receiving targeted assistance. Clarifying any remaining questions will help facilitate a better understanding of the concepts involved. Engaging actively in the problem-solving process is key to mastering the subject.
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...