One quarter sector is cut from a uniform circular disc of radius `R`. This sector has mass `M`. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. It moment of inertia about the axis of rotation is.
.

Options

A. `1/2MR^(2)`
B. `1/4MR^(2)`
C. `1/8MR^(2)`
D. `sqrt(2)MR^(2)`

Answer

A

Explanation

We cannot consider the quadrant as a single mass as the distance of different particles is different form the axis of rotation. So we take the help of calculus. Let us consider a segment as shown in figure. All masses lying in this segment are at a distance `r` form the axis and hence considered as a small differential mass `dm`. Let te thickness of the segment be `dr`.

The mass per unit of the quadrant `=M/(piR^(2)//4)=(4M)/(piR^(2))`
Area of the segment `=1/4[pi(r+dr)^(2)-pir^(2)]=1/4xx2pirdr=(pirdr)/2`
Mass of the segment `dm=(pirdr)/2xx(4M)/(piR^(2))=(2M)/R^(2)r^(2)rdr`
`MI` of the segment about `"nn"'=(2M)/R^(2)rdrxr^(2)(2M)/R^(2)r^(3)dr`
`MI` of the quadrant about `"nn"'=int_(0)^(R)(2M)/R^(2)r^(2)dr`
`(2M)/R^(2)xx(R^(4))/4xx(MR^(2))/2`