Question

The moment of inertia of a thin circular plate about its natural axis is $2.4\times {10}^{-2}kg{m}^2$. The moment of inertia of same circular plate about an axis in the plane of plate at a perpendicular distance half of radius of plate from its centre is:

• A. $0.8\times {10}^{-2}kg{m}^2$
• B. $1.2\times {10}^{-2}kg{m}^2$
• C. $0.6\times {10}^{-2}kg{m}^2$
• D. $2.4\times {10}^{-2}kg{m}^2$

Answer 1

The correct option is D $2.4\times {10}^{-2}kg{m}^2$

For a plate of mass $m$ and radius $r$, the MI about natural axis is $I_1=m{r}^2/2$

And MI about axis passing through center and in its plane is $m{r}^2/4$

Using parallel axis theorem, MI about an axis in the plane of plate at a perpendicular distance half of radius of plate from its center is $I_2=m{r}^2/4+m(r/2{)}^2=m{r}^2/2=I_1$

In the question, $I_1=2.4\times {10}^{-2}kg{m}^2$

So $I_2=2.4\times {10}^{-2}kg{m}^2$