Question

A uniform rod of length $1m$ is bent at ${90}^o$, so as to form two arms of equal of length. The centre of mass of this bent rod is :

• A. on the bisector of the angle $\left(\frac{1}{\sqrt{2}}\right)m$ from vertex
• B. on the bisector of the angle $\left(\frac{1}{2\sqrt{2}}\right)m$ from vertex
• C. on the bisector of the angle $\left(\frac{1}{2}\right)m$ from vertex
• D. on the bisector of the angle $\left(\frac{1}{4\sqrt{2}}\right)m$ from vertex

Answer 1

The correct option is D on the bisector of the angle $\left(\frac{1}{4\sqrt{2}}\right)m$ from vertex

After bending the rod of length $1m$ at ${90}^0$ it looks like,

In $△ABC$,

$\angle A+{90}^0$ and $AD$ is the bisector $LA$

$l\left(BC\right)=\sqrt{A{B}^2+A{C}^2}$

$=\sqrt{{\left(\frac{1}{4}\right)}^2+{\left(\frac{1}{4}\right)}^2}$

$l\left(BC\right)=\frac{1}{2\sqrt{2}}$

$l\left(BD\right)=l\left(DC\right)=l\left(AD\right)$.......(i)

$l\left(BC\right)=l\left(BD\right)+l\left(DC\right)$

$\therefore \frac{1}{2\sqrt{2}}=2l\left(OC\right)$

$\therefore l\left(DC\right)=\frac{1}{4\sqrt{2}}$

from eqn (i)

$l\left(OC\right)=l\left(AD\right)$

$\therefore l\left(AD\right)=\frac{1}{4\sqrt{2}}$

$\therefore$ Centre of mass is on ${}^{\prime }{D}^{\prime }$ on the bisector of the angle

$A$ at $\left(\frac{1}{4\sqrt{2}}\right)$ from vertex $A$.