Question

A uniform thin rod AB of length $L$ has linear mass density $\mu \left(x\right)=a+{\displaystyle \frac{bx}{L}}$​, where $x$ is measured from A. If the centre of mass of the rod lies at a distance of ${\displaystyle \frac{7L}{12}}$ from A, then a and b are related as
$\begin{array}{rl}& A.2a=b\\ & B.a=2b\\ & C.a=b\\ & D.3a=2b\end{array}$

Answer 1

Hint: Find out the centre of mass of the rod AB and substitute the terms mentioned in the question in it. Rearrange the equation and reach at the relation between a and b. The centre of mass is the point with respect to a body where the total mass of the body is assumed to be concentrated. This all will help you in answering this question.

Complete step by step answer:
The centre of mass of the rod is to be found. The length of the rod has been mentioned as $L$, the linear mass density has been given by the equation,
$\mu \left(x\right)=a+{\displaystyle \frac{bx}{L}}$
Where $x$ is measured from point A.
The distance of the centre of mass from the point A is given as,
$x_{cm}={\displaystyle \frac{7L}{12}}$
The centre of mass of the body is given as,
$x_{cm}={\displaystyle \frac{{\int }_0^L\mu \left(x\right)xdx}{{\int }_0^L\mu \left(x\right)dx}}$
Substituting the values in it will give,
$x_{cm}={\displaystyle \frac{{\int }_0^L(a+{\displaystyle \frac{bx}{L}})xdx}{{\int }_0^L(a+{\displaystyle \frac{bx}{L}})dx}}$
Performing the integration will be written as,
${\displaystyle \frac{7L}{12}}={\displaystyle \frac{a{\displaystyle \frac{L^2}{2}}+b{\displaystyle \frac{L^2}{3}}}{aL+{\displaystyle \frac{bL}{2}}}}$
We can cancel the length term as it is common in both the equations. That is we can write that,
${\displaystyle \frac{7}{12}}={\displaystyle \frac{{\displaystyle \frac{a}{2}}+{\displaystyle \frac{b}{3}}}{a+{\displaystyle \frac{b}{2}}}}={\displaystyle \frac{{\displaystyle \frac{3a+2b}{6}}}{{\displaystyle \frac{2a+b}{2}}}}={\displaystyle \frac{3a+2b}{3(2a+b)}}$
Simplifying the equation further can be written as,
${\displaystyle \frac{7}{4}}={\displaystyle \frac{3a+2b}{(2a+b)}}$
Now let us rearrange the equation. That is,
$14a+7b=12a+8b$
From this we can reach the final equation. That is,
$2a=b$

So, the correct answer is “Option A”.

Note: In the case of simple rigid bodies with a uniform density, the centre of mass is placed at the centroid. The centre of mass is the specific point at the centre of a distribution of mass in space that is having the characteristics that the weighted position vectors with respect to this point are resulting in being zero.