Question

What is the center of mass of the object below?
The shape is uniformly dense.
All answers are stated with respect to the origin, which is at the center of the large sphere.

• A. $(0,\frac{R}{2})$
• B. $(0,\frac{R}{4})$
• C. $(0,-\frac{R}{4})$
• D. $(0,-\frac{R}{6})$
• E. $(0,-\frac{R}{3})$

Answer 1

The correct option is D $(0,-\frac{R}{6})$

It is clear from the diagram, that both the bodies, circle of radius R, with hole and circle of radius R/2 are symmetrical to Y -axis, therefore their center of mass lies on Y-axis i.e. $x=0$.

This diagram can be considered as , two circular discs (of same material) are superimposed, one of mass m and radius R and other of mass -m' and radius R/2.

Now, surface density of disc of mass $m$,

$\rho =m/\pi R^2$,

surface density of disc of mass $m^{\prime }$,

$\rho =m^{\prime }/\pi (R/2{)}^2$,

hence, $m/\pi R^2=m^{\prime }/\pi (R/2{)}^2$,

or $m=4m^{\prime }$ ....................eq1,

If we consider the two discs as point masses,

therefore, center of mass of the given system,

$x=0$ ,

and $y=\frac{m\times 0-m^{\prime }\times R/2}{m-m^{\prime }}$,

or $y=\frac{-m^{\prime }\times R/2}{4m^{\prime }-m^{\prime }}=-R/6$,

center of mass of the given system is, $(0,-R/6)$