Another rotation problem -- Acceleration of a ring rolling on a surface

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The discussion centers on a physics problem involving a rotating ring on a surface, specifically analyzing the acceleration of the center, the frictional force from the ground, and the normal reaction when the ring's geometric center has a velocity of $$ \sqrt{gR} $$. Participants are encouraged to provide their attempts at solving the problem to facilitate guidance. The conversation emphasizes the importance of showing work to receive constructive feedback. The thread also references a related discussion for additional context.
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[Mentors' note - this question was spun off from https://www.physicsforums.com/threa...urface-but-with-a-twist.1078618/#post-7243282 which contains the homework template information]

I'm not sure if I can add on by adding a similar question, if I am not allowed someone please tell me.

A ring is rotating as shown in the figure. At that particular instant when the velocity of the geometric center is $$ \sqrt{gR} $$ find the
1. Acceleration of center
2. Frictional force from ground
3. Normal reaction
 

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palaphys said:
[Mentors' note - this question was spun off from https://www.physicsforums.com/threa...urface-but-with-a-twist.1078618/#post-7243282 which contains the homework template information]

I'm not sure if I can add on by adding a similar question, if I am not allowed someone please tell me.

A ring is rotating as shown in the figure. At that particular instant when the velocity of the geometric center is $$ \sqrt{gR} $$ find the
1. Acceleration of center
2. Frictional force from ground
3. Normal reaction
Where is your attempt?
 
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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