Question

A wireframe is made of a wire of uniform cross-section, which is shown in figure. ABC, HGF and DIE are semicircular arcs of radius r. CD=DO=OE=EF=r and 'O' is the centre of circle. Then centre of mass of frame is (mass per unit length is $\lambda$) :

• A. At distance $\left(\frac{2r}{3\pi +4}\right)$ towards left of O
• B. At distance $\left(\frac{2r}{3\pi +4}\right)$ towards right of O
• C. At distance $\left(\frac{4r}{3\pi +4}\right)$ towards left of O
• D. At distance $\left(\frac{4r}{3\pi +4}\right)$ towards left of O

Answer 1

The correct option is A At distance $\left(\frac{2r}{3\pi +4}\right)$ towards left of O

Using the formula of the position of centre of mass for the shown configuration

$X_{com}=\frac{4r\lambda \times 0-\pi r\lambda \times 2r/\pi +\pi r\lambda \times r-\pi r\lambda \times r}{3\pi r\lambda +4r\lambda }$
=$\left(\frac{-2r}{3\pi +4}\right)$
Therefore, COM is at a distance $\left(\frac{2r}{3\pi +4}\right)$ towards left of O.