Infinite rods of uniform mass density and len | vedclass

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Question

Let linear mass density be $\lambda$

$\mathrm{X}_{\mathrm{Cm}}=\frac{\sum \mathrm{m}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{m}_{\mathrm

Vedclass · Accepted Answer

$L/3$

Vedclass · Answer

Let linear mass density be $\lambda$

$\mathrm{X}_{\mathrm{Cm}}=\frac{\sum \mathrm{m}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{m}_{\mathrm{i}}}=\frac{(\lambda \mathrm{L})(\mathrm{L} / 2)+(\lambda \mathrm{L} / 2)(\mathrm{L} / 4)+\ldots}{\lambda \mathrm{L}+\lambda \mathrm{L} / 2 |+\ldots \ldots}$

$=\frac{\mathrm{L}}{2} \frac{\left(1+\frac{1}{4}+\ldots .\right)}{\left(1+\frac{1}{2}+\ldots .\right)}=\frac{\mathrm{L}}{2} \frac{\left(\frac{1}{1-\frac{1}{4}}\right)}{\left(\frac{1}{1-\frac{1}{2}}\right)}=\frac{\mathrm{L}}{3}$

Infinite rods of uniform mass density and length $L, L/2, L/4....$ are placed one upon another upto infinite as shown in figure. Find the $x-$ coordinate of centre of mass

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