Question

A small sector is cut from a uniform circular disc of radius $1\text{m}$. The sector subtends an angle ${90}^{\circ }$ at the center of the disc. This sector has mass $0.16\text{kg}$ and it is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is,

• A. $0.15{\text{kg m}}^2$
• B. $0.04{\text{kg m}}^2$
• C. $0.10{\text{kg m}}^2$
• D. $0.08{\text{kg m}}^2$

Answer 1

The correct option is D $0.08{\text{kg m}}^2$


Let $\sigma$ be the surface mass density of the given sector.

$\sigma =\frac{M}{\left(\frac{\pi R^2}{4}\right)}=\frac{4M}{\pi R^2}$

$\sigma =\frac{4\left(0.16\right)}{\pi \times (1{)}^2}=\frac{0.64}{\pi }{\text{kg/m}}^2$

Let us consider an elemental strip of radius $x$ and thickness $dx$ as shown in figure.

Let the mass of the strip be $dm$.

$\therefore dm=\sigma dA=\sigma \times \frac{2\pi xdx}{4}$

$dm=\frac{0.64}{\pi }\times \frac{\pi }{2}xdx[\because \text{It is quarter circle}]$

$dm=0.32xdx$

Moment of inertia of the thin strip is,

$dI=dm{x}^2$

Integrating both sides with proper limit we get,

${\int }_0^{I}dI={\int }_0^{1}dm{x}^2$

$I={\int }_0^{1}0.32x^{3}dx=0.32{\left[\frac{x^4}{4}\right]}_0^1=\frac{0.32}{4}$

$\therefore I=0.08{\text{kg m}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.

Alternate solution: If an object can be obtained by repeating a certain unit then the moment of inertia of the object and the repeating units resembles about the same axis. $\therefore$ Moment of inertia of the disc = Moment of inertia of given sector $I=\frac{M{R}^2}{2}=\frac{0.16\times (1{)}^2}{2}=0.08{\text{kg m}}^2$