Question

A non-uniform rod having mass per unit length as $\mu =ax$ (a is constant). If its total mass is M and length L, the centre of mass is at :

• A. $x=\left(3/4\right)L$
• B. $x=\left(2/3\right)L$
• C. $x=\left(2/5\right)L$
• D. $x=\left(1/3\right)L$

Answer 1

The correct option is B $x=\left(2/3\right)L$

Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Choose an infinitesimal mass element dm located a distance x'. Let the length of the mass element be dx'.
Thus
$dm=\mu \left(x^{\prime }\right)d{x}^{\prime }$
The total mass is found by integrating the mass element over the length of the rod
$M={\int }_0^L\mu \left(x^{\prime }\right)d{x}^{\prime }=a{\int }_0^L{x}^{\prime }d{x}^{\prime }=\frac{a}{2}{x}^{\prime 2}{\mid }_0^L=\frac{a}{2}{L}^2$
or
$a=\frac{2M}{L^2}$
Now center of mass is calculated as
$x_{cm}=\frac{1}{M}{\int }_{body}xdm=\frac{1}{M}{\int }_0^L\mu \left(x^{\prime }\right)x^{\prime }d{x}^{\prime }=\frac{a}{M}{\int }_0^L{x}^{\prime 2}d{x}^{\prime }$
substituting the value of a
$\frac{2}{L^2}{\int }_0^L{x}^{\prime 2}d{x}^{\prime }=\frac{2}{3L^2}{x}^{\prime 3}{\mid }_0^L=\frac{2}{3L^2}(L^3-0)=\frac{2}{3}L$