Q. A circular disc of radius $R$ is removed from a bigger circular disc of radius 2$R$, such that the circumference of the discs coincide. The center of mass of the new disc is $\frac{\alpha}{R}\ $from the center of the bigger disc the value of $\alpha$ is
A) $\frac{1}{3}$
B) $\frac{1}{2}$
C) $\frac{1}{6}$
D) $\frac{1}{4}$
Explanation
(a) In this question distance of centre of mass
of new disc from the centre of mass of
remaining disc is $\alpha R$.
Mass of remaining disc
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$= M - \frac{M}{4} = \frac{3M}{4}$
$\therefore$ $- \frac{3M}{4}\alpha R + \frac{M}{4}R = 0$
$\therefore$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha = \frac{1}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.