Finding COM of L-Shaped Rod and Ball System

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The discussion focuses on finding the center of mass (CoM) of an L-shaped rod and a ball system. Participants suggest treating the L-shaped rod as two separate rods to simplify the calculation of the CoM. There is a request for more details about the homework assignment, including dimensions and mass values. The importance of seeing the entire problem is emphasized, as it may not always be necessary to find the CoM. The conversation highlights the need for clarity in physics problems to determine the best approach.
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Homework Statement
its for my physics hw
Relevant Equations
h
I suspect we can treat the L shaped rod as two rods and then find the com for that system and then find the com of the ball and that system
 
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Yes!

or you could find the com of one of the rods and the ball, and then find the com of the 1st system and the other rod,

or better ...
 
March said:
Homework Statement:: its for my physics hw
Relevant Equations:: h

I suspect we can treat the L shaped rod as two rods and then find the com for that system and then find the com of the ball and that system
Welcome @March !
Could you post the full homework and your attempt to solve it?
Have you been given any dimmensions and values for each mass?
 
March said:
Homework Statement:: its for my physics hw
Relevant Equations:: h

I suspect we can treat the L shaped rod as two rods and then find the com for that system and then find the com of the ball and that system
It is often unnecessary to find a CoM. We need to see the whole question.
 
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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