Centre of mass of non-uniform rod

AI Thread Summary
The discussion revolves around calculating the center of mass of a non-uniform rod with a linear increase in mass from the lighter end. The user attempts to define linear density using the equation λ = kx + a and integrates to find the center of mass but struggles to eliminate the constant k. They receive confirmation that their approach is correct but need guidance on simplifying the equation. Additionally, there is a reminder about proper posting etiquette, emphasizing the importance of typing out work instead of submitting images. The conversation highlights the challenges of integrating variable density functions in physics problems.
carlyn medona

Homework Statement


The mass of non uniform rod increases linearly with distance from lighter end. If m is mass of the rod and l it's total length a the linear density at lighter end, then found the distance of centre of mass from lighter end

Homework Equations


I put λ= kx+a where lands is linear density k a constant and x the distance from lighter end and integrated but can't get the answer in terms of m l and a

The Attempt at a Solution

 
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Please show us what you did.
 
So how do I eliminate k from the equation
 

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Remember that m = ∫(kx+a)dx
 
I think I made a mistake in my first calculation , is this one right?
 

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carlyn medona said:
I think I made a mistake in my first calculation , is this one right?
Yes this looks right.
 
Thanks for checking
 
carlyn medona said:
So how do I eliminate k from the equation

For future reference: you should not post images of handwritten work; actually take the time and trouble to type things out. Most helpers will not even look at all at such posted images, but you were lucky in this case to get a sympathetic reader. Please consult the post "Guidelines for students and helpers" by Vela, pinned to the start of this forum's file list.
 
Oops sorry for that.
 
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