The linear mass density (λ) of a rod of length L kept along x-axis varies as λ=α+β x; where α and β are positive constants. The centre of mass of the rod is at

Question

The linear mass density (λ) of a rod of length L kept along x-axis varies as λ=α+β x; where α and β are positive constants. The centre of mass of the rod is at (A) ((2 β+3 α L) L/2(2 β+α L)) (B) ((3 α+2 β L) L/3(2 α+β L)) (C) ((3 β+2 α L) L/3(2 β+α L)) (D) ((3 β+2 α L) L/3 β+2 α)

Tardigrade Admin · Accepted Answer

λ=α+β x d m=(α+β x) d x xc m=(∫ limits0L x(α+β x) d x/∫ limits0L(α+β x) d x) =(α ∫ limits0L x d x+β ∫ limits0L x2 d x/α ∫ limits0L d x+β ∫ limits0L x d x) xc m=((α L2/2)+(β L3/3)/α L+(β L2)2)

Q. The linear mass density $(λ)$ of a rod of length $L$ kept along $x$ -axis varies as $λ = α + β x$ ; where $α$ and $β$ are positive constants. The centre of mass of the rod is at

$\frac{( 2 β + 3 αL ) L}{2 ( 2 β + αL )}$

$\frac{( 3 α + 2 β L ) L}{3 ( 2 α + β L )}$

$\frac{( 3 β + 2 αL ) L}{3 ( 2 β + αL )}$

$\frac{( 3 β + 2 αL ) L}{3 β + 2 α}$

$λ = α + β x$
$d m = (α + β x) d x$
$x_{c m} = \frac{0 \int L x ( α + β x ) d x}{0 \int L ( α + β x ) d x}$
$= \frac{α 0 \int L x d x + β 0 \int L x ^{2} d x}{α 0 \int L d x + β 0 \int L x d x}$
$x_{c m} = \frac{\frac{α L ^{2}}{2} + \frac{β L ^{3}}{3}}{αL + \frac{β L ^{2}}{2}}$

Q. The linear mass density (λ) of a rod of length L kept along x-axis varies as λ=α+βx; where α and β are positive constants. The centre of mass of the rod is at

2(2β+αL)(2β+3αL)L​

3(2α+βL)(3α+2βL)L​

3(2β+αL)(3β+2αL)L​

3β+2α(3β+2αL)L​