Question

You are given a rod of length 'L'. The linear mass density varies as ${\lambda }^{\prime }$. $\lambda =a+bx$. Here a & b are constants and the mass of the rod increases as x increases. Find the mass of the rod.

• A. $a{L}^2$
• B. $aL+\frac{b{L}^2}{2}$
• C. $\frac{L^2(a+b)}{2}$
• D. None of these

Answer 1

The correct option is B

$aL+\frac{b{L}^2}{2}$

We cannot calculate the mass of the rod by simply multiplying linear mass density '${}^{\prime }{\lambda }^{\prime }$ will the length of the rod as ${}^{\prime }{\lambda }^{\prime }$ is not constant.

Hence we have to divide entire rod length into a number of elements and add the mass of all elements.

$M=d{m}_1+d{m}_2+d{m}_3+.......$

This step can be written in terms of integration.

$M=\int dm$

If we take an element dx on the rod at a distance x from the left end of the rod.

Mass of this element $dm=\lambda .dx$

$dm=(a+bx)dx$

Hence total mass $M=\int dm\Rightarrow M=\underset{0}{\overset{L}{\int }}(a+bx)dx$

$M=\underset{0}{\overset{L}{\int }}adx+\underset{0}{\overset{L}{\int }}bxdx=a\underset{0}{\overset{L}{\int }}dx+b\underset{0}{\overset{L}{\int }}xdx=a[x{]}_0^L+b{\left[\frac{x^2}{2}\right]}_0^L\Rightarrow M=aL+\frac{b{L}^2}{2}$