If the linear mass density of a rod of length L lying along x-axis and origin at one end varies as -A-Bx- where A and B are constants- find the coordinates of the centre of mass-

As rod is kept along x axisHence, Y _ cm =0, Z _ cm =0 For coordinate x _ cm =∫ _ 0 ^ L x dm ∫ _ 0 ^ L L #160; dm=λ .dx=(A+Bx)dx #160; ...

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Standard XII

Physics

Young's Modulus of Elasticity

If the linear...

Question

If the linear mass density of a rod of length $L$ lying along x-axis and origin at one end varies as $λ = A + B x$ , where $A$ and $B$ are constants, find the coordinates of the centre of mass.

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Solution

As rod is kept along x-axis
Hence, $Y_{c m} = 0, Z_{c m} = 0$
For coordinate
$x_{c m} = \frac{\int_{0}^{L} x d m}{\int_{0}^{L} L} d m = λ . d x = (A + B x) d x$
$x_{c m} = \frac{\int_{0}^{L} x (A x + B) d x}{\int_{0}^{L} (A + B x) d x} = \frac{\frac{A L^{2}}{2} + \frac{B L^{2}}{3}}{A L + \frac{B L^{2}}{2}} = \frac{L (3 A + 2 B L)}{3 (2 A + B L)}$
Here coordinates of the centre of mass are $(\frac{L (3 A + 2 B L)}{3 (2 A + B L)}, 0, 0)$

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If the linear mass density of a rod of length L lying along x-axis and origin at one end varies as λ=A+Bx, where A and B are constants, find the coordinates of the centre of mass.

As rod is kept along x-axisHence, Ycm=0,Zcm=0For coordinatexcm=∫L0xdm∫L0Ldm=λ.dx=(A+Bx)dx xcm=∫L0x(Ax+B)dx∫L0(A+Bx)dx=AL22+BL23AL+BL22=L(3A+2BL)3(2A+BL)Here coordinates of the centre of mass are (L(3A+2BL)3(2A+BL),0,0)

If the linear mass density of a rod of length $L$ lying along x-axis and origin at one end varies as $λ = A + B x$ , where $A$ and $B$ are constants, find the coordinates of the centre of mass.