Question

A uniform rod of length $10\text{m}$ is placed along $x$ axis as shown in figure. The distance of the centre of mass of the rod from the origin is

• A. $20\text{m}$
• B. $5\text{m}$
• C. $15\text{m}$
• D. $25\text{m}$

Answer 1

The correct option is C $15\text{m}$

Let the mass of uniform rod be $M$.
Mass per unit length of the rod $\left(\lambda \right)=\frac{M}{L}=$ constant
($\because$ Rod is uniform)
Let $dm$ be the mass of elemental length $dx$ at a distance $x$ from the origin.
$\therefore dm=\left(\frac{M}{L}\right)dx=\left(\frac{M}{10}\right)dx$
To find position of centre of mass,
$x_{COM}=\frac{\int xdm}{\int dm}=\frac{\underset{10}{\overset{20}{\int }}x\times \left[\frac{M}{10}dx\right]}{\underset{10}{\overset{20}{\int }}\frac{M}{10}dx}$
$=\frac{\underset{10}{\overset{20}{\int }}xdx}{\underset{10}{\overset{20}{\int }}dx}=\frac{1}{10}{\left[\frac{x^2}{2}\right]}_{10}^{20}$
$=\frac{1}{10}\left(\frac{(20{)}^2-(10{)}^2}{2}\right)$
$=15\text{m}$