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Question: A thin bar of length \(L\) has a mass per unit length \(\lambda \), that increases linearly with dis...

A thin bar of length LL has a mass per unit length λ\lambda , that increases linearly with distance from one end. If its total mass be MM and its mass per unit length at the lighter end be λ0{{\lambda }_{0}}, then what will be the distance of the centre of mass from the lighter end?
A.L2−λ0L24M B.L3+λ0L24M C.L3+λ0L28M D.2L3−λ0L26M \begin{aligned} & A.\dfrac{L}{2}-\dfrac{{{\lambda }_{0}}{{L}^{2}}}{4M} \\\ & B.\dfrac{L}{3}+\dfrac{{{\lambda }_{0}}{{L}^{2}}}{4M} \\\ & C.\dfrac{L}{3}+\dfrac{{{\lambda }_{0}}{{L}^{2}}}{8M} \\\ & D.\dfrac{2L}{3}-\dfrac{{{\lambda }_{0}}{{L}^{2}}}{6M} \\\ \end{aligned}

Explanation

Solution

First of all write the equation for the mass per unit length of the bar. Rearrange this equation in terms of acceleration. Centre of mass of a body is the imaginary point we are assuming where the whole mass of the body has been concentrated. Then find the centre of mass equation. Substitute the needed parameters in it and simplify. This will help you in answering this question.

Complete answer:
Let the bar be placed along the x axis with its lighter end kept at the origin.

We can write the mass per unit length as,
λ(x)dx=[λ0x+ax22]0L\lambda \left( x \right)dx=\left[ {{\lambda }_{0}}x+\dfrac{a{{x}^{2}}}{2} \right]_{0}^{L}
Let us apply the limits in it,
λ(x)dx=λ0L+aL22\lambda \left( x \right)dx={{\lambda }_{0}}L+\dfrac{a{{L}^{2}}}{2}
Rearranging this equation will be written as,
a=2(M−λ0L)L2a=\dfrac{2\left( M-{{\lambda }_{0}}L \right)}{{{L}^{2}}}
Let us assume that xˉ\bar{x} be the centre of mass.
Therefore the centre of mass of the body will be given as,
xˉ=∫xλ(x)dxM\bar{x}=\dfrac{\int x\lambda \left( x \right)dx}{M}
Substituting the values in it will give,
xˉ=[λ0x22+ax33]0LM\bar{x}=\dfrac{\left[ {{\lambda }_{0}}\dfrac{{{x}^{2}}}{2}+a\dfrac{{{x}^{3}}}{3} \right]_{0}^{L}}{M}
Let us apply the limits in this equation which can be shown as,
xˉ=λ0L22+aL33M\bar{x}=\dfrac{{{\lambda }_{0}}\dfrac{{{L}^{2}}}{2}+a\dfrac{{{L}^{3}}}{3}}{M}
Now let us substitute aa in this equation,
xˉ=λ0L22+2(M−λ0L)L3M\bar{x}=\dfrac{{{\lambda }_{0}}\dfrac{{{L}^{2}}}{2}+\dfrac{2\left( M-{{\lambda }_{0}}L \right)L}{3}}{M}
From this after simplifying the equation, we can write that,
xˉ=2L3−λ0L26M\bar{x}=\dfrac{2L}{3}-\dfrac{{{\lambda }_{0}}{{L}^{2}}}{6M}
This will be the distance of the centre of mass from the lighter end.

The answer has been mentioned as option D.

Note:
The centre of mass is defined as a location described relative to a body or system of objects. It will be the average location of all the parts of the system, weighted with respect to their masses. For rigid bodies with uniform density, the centre of mass is found to be at the centroid.