Find the center of mass of an arc

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The discussion centers on finding the center of mass of an arc shaped like the letter "B." One participant initially approached the problem by treating it as a semicircle, yielding an approximate answer. Another contributor pointed out that this method overlooked the mass of the upright line in the "B," which significantly affects the calculation. The conversation highlights the importance of considering all components of the shape to achieve an accurate center of mass. Multiple methods exist for calculating the center of mass, emphasizing the need for clarity in visual aids for better understanding.
Istiak
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Homework Statement
Find center of mass of a arc (letter B)
Relevant Equations
x_com = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2}
1625387927009.png


In my mind, I had cut half of B and, thought it's semi-circle. Then, I was trying to find Center of Mass by taking it as semi-circle. But, I get an answer which is approximately, close to main answer. Someone else had solved it another way

1625388180302.png
This way I can get the accurate answer. But, the picture is very blurry that's why I can't understand it. Why both answers aren't same? Is there any other way to calculate Center of Mass?
 
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Istiakshovon said:
Homework Statement:: Find center of mass of a arc (letter B)
Relevant Equations:: x_com = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2}

View attachment 285438

In my mind, I had cut half of B and, thought it's semi-circle. Then, I was trying to find Center of Mass by taking it as semi-circle. But, I get an answer which is approximately, close to main answer. Someone else had solved it another way

View attachment 285439This way I can get the accurate answer. But, the picture is very blurry that's why I can't understand it. Why both answers aren't same? Is there any other way to calculate Center of Mass?
Your solution seems to have ignored the mass of the upright line in "B".
 
haruspex said:
Your solution seems to have ignored the mass of the upright line in "B".
Which solution? I had attached two pictures. One is done by me (my friend) and, another had done by another person in Internet..
 
Yes your solution (first picture in the OP) doesn't take into account the "base line" of "B", that is the segment that consists of the diameters of the two semicircles and has total length 4r.
 
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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