Question

Consider a thin stick of length L. standing on one of its ends on a friction less surface. It is slightly pushed at the other end of the rod. Then path of centre of mass of the rod is

• A. $x=0$
• B. $x^2+y^2=\frac{L^2}{4}$
• C. $y=0$
• D. $\frac{x^2}{L^2/4}+\frac{y^2}{L^2}=1$

Answer 1

The correct option is B $x^2+y^2=\frac{L^2}{4}$

Given,

stick is very thin, Length of rod is L

From figure, stick one end is placed at center of axis $(x=0,y=0)$

Center of mass of stick is at $\frac{L}{2}$ distance from edge.

Center of mass of stick, follow circular path. With radius $r=\frac{L}{2}$ center $(a=0,b=0)$

Apply equation of circle.

${(x-a)}^2+{(y-b)}^2=r^2$

${(x-0)}^2+{(y-0)}^2={\left(\frac{L}{2}\right)}^2$

$x^2+y^2={\frac{L}{4}}^2$

Path followed by center of mass of stick $x^2+y^2={\frac{L}{4}}^2$