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A rod of length L has non uniformly distributed mass along its length.Its linear mass density lamda (x)= M0 /L2(L+X) M0 is a constant and x is the from one end of the rod.Position of CM =K/9 find the value of K

kanchana Rao , 11 Years ago
Grade
anser 1 Answers
Sumit Majumdar

Last Activity: 10 Years ago

Dear student,
Can you confirm that the linear mass denisty is given by:
\lambda= \frac{M_{0}}{L^{2}}(L+x)
Hence the centre of mass would be given by:
\Text{Centre of mass}(C)={\int _{0}^{L} \lambda dx}=\frac{M_{0}}{L^{2}} \int_{0}^{L} \left ( L+x \right )dx=\frac{M_{0}}{L^{2}} \left ( L^2+\frac{L^2}{2} \right )=\frac{3}{2}M_0
Hence,
K=\frac{27}{2}M_0
Regards

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