Question

Two small spheres of masses 10 kg and 30 kg are joined by a rod of length 0.5 m and of negligible mass. The M.I. of the system about a normal axis through centre of mass of the system is

• A. 1.875 $kg{m}^2$
• B. 2.45 $kg{m}^2$
• C. 0.75 $kg{m}^2$
• D. 1.75 $kg{m}^2$

Answer 1

The correct option is A 1.875 $kg{m}^2$

$Small$ spheres so they can be treated as $point$ mass.

The center of mass will be at a distance $x=0.5meter\times \frac{30}{10+30}=\frac{3}{8}meter$ from the $10kg$ sphere i.e. at a distance $X=\frac{1}{2}-\frac{3}{8}=\frac{1}{8}meter$ from $30kg$ sphere.

Now the rod being massles will not contribute in the M.I,

so the M.I will be $I=m_1{r}_1^2+m_2{r}_2^2=10kg\times \frac{9}{64}{m}^2+30kg\times \frac{1}{64}{m}^2=\frac{15}{8}Kg{m}^2=1.875Kg{m}^2$