I have to calculate the center of mass (COM) of a rod having a non-linear mass density, such that \( \dfrac{dm}{dx} = x\). While I was calculating the COM, my friend interrupted me and challenged my answer, which left me a little bit confused.

See if you can help me out:

**Case 1**: (The way I solved it) Since \( \dfrac{dm}{dx} = x\), I used the formula to calculate it such that \( \dfrac{ \int_0^1 x \, dm }{ \int_0^m \, dm } \) and then put the values of \(dm\), I get COM as \( \dfrac{\int_0^l x \, dx \cdot x }{\int_0^l x \, dx } = \dfrac{2l^3}{3l^2} = \dfrac{2l}3 \).

**Case 2**: (The way my friend solve it) His definition is COM lies at the spot such that the masses are equally divided in the 2 portions of the rod. He did: \( \displaystyle \int_0^l x \, dx \ cdot x = \dfrac{x^2}2 \). \( \dfrac{x^2}2 = \dfrac{l^2}2 - \dfrac{x^2}2 \) after using proper limits. He got COM at \(\dfrac l{\sqrt2} \).

Please tell me who is right and who is wrong with proper reasons.

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## Comments

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TopNewestPlease refrain from using all caps while typing. Its a bad habit on internet

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what do you mean by " caps " and what reason should i give him so that he is really convinced as i already told him that his definition is wrong but still argued with me( plz give a reason so that he is spell bound), thanks for reassurance.

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writing all words with capslock is a bad habit bro!

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You are correct and your friend is wrong .That isn't the definition of centre of mass

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Centre of mass of some bodies even lie outside the object!.

For example consider a V Shaped uniform rod . Its com will lie somewhere on the axis of symmetry rather than on the body.

Saying that masses are equally distributed on both sides isn't a technically correct thing i suppose.

Centre of mass is Basically the weighted mean of small mass elements as a function of length or so.

2l/3 in your answer signifies that it is the length where whole mass of the body can be concentrated or viz that is the final mean position of all the masses.

Similarly consider a Half disc whose COM Lies at a height 4R/3pi from centre of disc . Is mass distribution equal above and below the centre of mass ?

obviously not they would have different areas

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Your friend has some nice ideas - however the CoM lies at the point at which the moments of the masses on each side are equal, not the masses themselves.

If the rod lies from \(x = 0\) to \(x = L\), and its centre of mass is at \(x = x_0\), then, \[ \int\limits_{0}^{x_0} |x - x_0| \cdot dm = \int\limits_{x_0}^{L} |x - x_0| \cdot dm \]

It's really easy to prove, from the identity \[ \int\limits_{0}^{L} (x - x_0) dm = 0 \]

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