Question

You are supp;ied with three identical rods of same length and mass. if the length of each rods is $2\pi$ two of them are converted into rings and then placed over the third rod as origin of the coordinate system the cordinate of the centre of mass will be ( you may assume AB as x-axis of the cordinate system)

• A. $(\frac{\pi }{2},\frac{1}{3})$
• B. $(\frac{\pi }{2},\frac{2}{3})$
• C. $(\pi ,\frac{1}{3})$
• D. $(\pi ,\frac{2}{3})$

Answer 1

The correct option is D $(\pi ,\frac{2}{3})$

As all the rods have $same$ length so they should have $same$ mass.

Now the center of mass of the circular rods will be at $their$ respective $centers$ i.e just above the corners of the straight

rod.

As the circular rod have circumference as $2\pi$ so the radius will be $r=\frac{circumference}{2\pi }=\frac{2\pi }{2\pi }=1$

so taking the left end as the origin and the straight rod as X-axis the center of mass of straight tod will be

$\left(\frac{2\pi }{2}\right(middle),0)$ or $(\pi ,0)$

and that of left ring at $(0,1(\text{vertical radius)})$

and that of right ring at $(2\pi ,1)$

so the center of the system will be at $(x,y)$ given by $x=\frac{\Sigma mx}{m+m+m}=\frac{1}{3}(\pi +0+2\pi )=\pi$

similarly $y=\frac{1}{3}(0+1+1)=\frac{2}{3}$