Question

About which of the following axes moment of inertia of a circular disc is minimum?

• A. Axis passes through its center and perpendicular to its plane.
• B. Axis along the diameter.
• C. Axis along the tangent and in its own plane
• D. Axis along the tangent and perpendicular to its plane.

Answer 1

The correct option is B

Axis along the diameter.

Explanation for the correct option:

Option (B)

1. The moment of inertia of a body having a uniform pure rotation about an axis in space is defined as $\sum _i{m}_i{r_i}^2$ is the mass of the I th particle of the body and $r_i$ its distance from the axis concerned.
2. The moment of inertia of a circular disc, when the axis of rotation passes through the diameter is $\frac{m{r}^2}{4}$.

Explanation for the incorrect options:

Option (A)

1. The moment of inertia of a circular disc, when the axis of rotation passes through the center of the disc and perpendicular to its plane is $\frac{1}{2}m{r}^2$, where m and r are the mass and radius of the circular disc.

Option (C)

1. The moment of inertia of a circular disc, when the axis of rotation passes through the tangent and parallel to the plane is $\frac{5m{r}^2}{4}$

Option (D)

1. The moment of inertia of a circular disc, when the axis of rotation passes through the tangent and perpendicular to the plane is $\frac{3}{2}m{r}^2$

So, the descending order of moment of inertia is $\frac{3}{2}m{r}^2$ $>$ $\frac{5m{r}^2}{4}$$>$ $\frac{1}{2}m{r}^2$ $>$ $\frac{m{r}^2}{4}$.

So, option (B) is correct.