Question

A uniform rod of length one meter is bent at its midpoint to make ${90}^0$. The distance of centre of mass from the centre of rod is(in cm)?

• A. $20.2$
• B. $13.4$
• C. $15$
• D. $35.36$

Answer 1

The correct option is A $20.2$

Let mass of total rod $=2m$

Mass of half rod $=m$

Coordinate of center of mass after bending separately $(E25,0)$ and $(50,-25)$

Center of mass coordinate $(x,y)$

$x=\frac{25m+50m}{2m}=\frac{75}{2}cmy=\frac{0\times m-25\times m}{2m}=-\frac{25}{2}cm$

Distance from center of rod $A(50,0)$ is $d$

$d=\sqrt{(x-50{)}^2+(y-0{)}^2}=\sqrt{{(50-\frac{75}{2})}^2+{(-\frac{25}{2})}^2}=\sqrt{{\left(\frac{25}{2}\right)}^2+{\left(\frac{25}{2}\right)}^2}=\frac{25}{2}\sqrt{2}=\frac{25}{\sqrt{2}}cm=20.2cm$

Option : A