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Question: The linear mass density of a ladder of length \(l\) increases uniformly from one end \(A\) to the ot...

The linear mass density of a ladder of length ll increases uniformly from one end AA to the other end BB ,
(i) Form an expression for linear mass density as a function of distance xx from end AA where linear mass density λ0{\lambda _0} . The density at one end being twice that of the other end.
(ii) Find the position of the center of mass from end AA .

Explanation

Solution

Use the formula of the linear mass density and find it at the point AA itself by substituting the distance and the density to obtain the expression. Then substitute the obtained expression in the value of the distance of the center of mass to obtain the answer.

Formula used:
The formula of the linear mass density is given by
λ=Ax+B\lambda = Ax + B
Where λ\lambda is the linear mass density and AandBA\,and\,B are the two ends of the ladder.

Complete step by step solution:
It is given that the
Length of the ladder is ll
The linear mass density is λ0{\lambda _0}
At the point AA , the distance between the points is zero and the linear mass density λ=λ0\lambda = {\lambda _0}. And at point BB , at a distance xx from the point AA , The density at one end is twice the other end, hence λ=2λ0\lambda = 2{\lambda _0} .
From these boundary points, we get A=λ0lA = \dfrac{{{\lambda _0}}}{l} and the value of the B=λ0B = {\lambda _0} . Substituting these in the formula of the linear mass density,
λ=Ax+B\lambda = Ax + B
λ=λ0lx+λ0\lambda = \dfrac{{{\lambda _0}}}{l}x + {\lambda _0}
It is known that the
X=01xdm01dmX = \dfrac{{\int {_0^1xdm} }}{{\int {_0^1dm} }} and the dm=λdxdm = \lambda dx
Substitute the value of the λ\lambda in the dmdm formula,
Hence the dmdm obtained is λ0xl+λ0\dfrac{{{\lambda _0}x}}{l} + {\lambda _0} .
Substituting the value of the dmdm in the formula of XX .
X=01x(λ0xl+λ0)dx01(λ0xl+λ0)dxX = \dfrac{{\int {_0^1x\left( {\dfrac{{{\lambda _0}x}}{l} + {\lambda _0}} \right)} dx}}{{\int {_0^1\left( {\dfrac{{{\lambda _0}x}}{l} + {\lambda _0}} \right)} dx}}
By simplifying the above step, we get
X=519X = \dfrac{{51}}{9}
Hence the centre of mass is at a distance of 519cm\dfrac{{51}}{9}\,cm from the point AA .

Note: The integration is done to find the distance of the center of the mass from the given point. Remember the integration formula of the distance and also the linear mass density. The linear mass density is the mass per unit length.