Question

Two point masses $m$ and $3m$ are placed at distance. The moment of inertia of the system about an axis passing through the center of mass of system and perpendicular to the line joining the point masses is

• A. $\frac{3}{5}m{r}^2$
• B. $\frac{3}{4}m{r}^2$
• C. $\frac{3}{2}m{r}^2$
• D. $\frac{6}{7}m{r}^2$

Answer 1

The correct option is B $\frac{3}{4}m{r}^2$

Suppose the separation is $r$ then the center of mass from the mass,$m_1$ will be at a distance given by $r_1=\frac{m_2}{m_1+m_2}r$

And the center of mass from the mass,$m_2$ will be at a distance given by $r_2=\frac{m_1}{m_1+m_2}r$

So the moment of inertia will be $I=I_1+I_2=m_1{r}_1^2+m_2{r}_2^2=\frac{m_1{m}_2^2}{(m_1+m_2{)}^2}{r}^2+\frac{m_2{m}_1^2}{(m_1+m_2{)}^2}{r}^2$

Putting $m_1=m$ and $m_2=3m$ we get $I=\frac{3m{r}^2}{4}$

Option B is correct.