A uniform circular disk of radius $R$ begins with a mass $M$; about an axis through the center of the disk and perpendicular to the plane of the disk the moment of inertia is $I_\text{0}=\frac{1}{2}MR^2$. A hole is cut in the disk as shown in the diagram. In terms of the radius $R$ and the mass $M$ of the original disk, what is the moment of inertia of the resulting object about the axis shown?

[asy]
size(14cm);
pair O=origin;
pair
A=O,
B=(3,0),
C=(6,0);

real
r_1=1,
r_2=.5;


pen my_fill_pen_1=gray(.8);
pen my_fill_pen_2=white;
pen my_fill_pen_3=gray(.7);
pen my_circleline_draw_pen=black+1.5bp;

//fill();

filldraw(circle(A,r_1),my_fill_pen_1,my_circleline_draw_pen);
filldraw(circle(B,r_1),my_fill_pen_1,my_circleline_draw_pen);

// Ellipse
filldraw(yscale(.2)*circle(C,r_1),my_fill_pen_1,my_circleline_draw_pen);


draw((C.x,C.y-.75)--(C.x,C.y-.2), dashed);
draw(C--(C.x,C.y+1),dashed);
label("axis of rotation",(C.x,C.y-.75),3*S);

// small ellipse
pair center_small_ellipse;
center_small_ellipse=midpoint(C--(C.x+r_1,C.y));
//dot(center_small_ellipse);
filldraw(yscale(.15)*circle(center_small_ellipse,r_1/2),white);


pair center_elliptic_arc_arrow;
real gr=(sqrt(5)-1)/2;
center_elliptic_arc_arrow=(C.x,C.y+gr);
//dot(center_elliptic_arc_arrow);

draw(//shift((0*center_elliptic_arc_arrow.x,center_elliptic_arc_arrow.y-.2))*
(
yscale(.2)*
(
arc((center_elliptic_arc_arrow.x,center_elliptic_arc_arrow.y+2.4), .4,120,360+60))
),Arrow);
//dot(center_elliptic_arc_arrow);

// lower_Half-Ellipse
real downshift=1;

pair C_prime=(C.x,C.y-downshift);
path lower_Half_Ellipse=yscale(.2)*arc(C_prime,r_1,180,360);
path upper_Half_Ellipse=yscale(.2)*arc(C,r_1,180,360);
draw(lower_Half_Ellipse,my_circleline_draw_pen);
//draw(upper_Half_Ellipse,red);



// Why here ".2*downshift" instead of downshift seems to be not absolutely clean.
filldraw(upper_Half_Ellipse--(C.x+r_1,C.y-.2*downshift)--reverse(lower_Half_Ellipse)--cycle,gray);

//filldraw(shift(C-.1)*(circle((B+.5),.5)),my_fill_pen_2);//
filldraw(circle((B+.5),.5),my_fill_pen_2);//shift(C-.1)*

/*
filldraw(//shift((C.x,C.y-.45))*
yscale(.2)*circle((C.x,C.y-1),r_1),my_fill_pen_3,my_circleline_draw_pen);
*/


draw("$R$",A--dir(240),Arrow);
draw("$R$",B--shift(B)*dir(240),Arrow);

draw(scale(1)*"$\scriptstyle R/2$",(B+.5)--(B+1),.5*LeftSide,Arrow);
draw(scale(1)*"$\scriptstyle R/2$",(B+.5)--(B+1),.5*LeftSide,Arrow);

[/asy]
(A) $\text{(15/32)}MR^2$
(B) $\text{(13/32)}MR^2$
(C) $\text{(3/8)}MR^2$
(D) $\text{(9/32)}MR^2$
(E) $\text{(15/16)}MR^2$
1. Let \( I_h \) and \( I \) denote the moments of inertia of the hole piece that was cut out and the final moment of inertia, respectively. Then, it is obvious that 
   \[
   I_h + I = I_0.
   \]
   Given that \( I_0 \) is the moment of inertia of the original disk, we have:
   \[
   I_0 = \frac{1}{2}MR^2.
   \]
   Therefore, we need to find \( I_h \).

2. The hole cut out is a smaller disk of radius \( \frac{R}{2} \) and mass \( \frac{M}{4} \) (since the mass is proportional to the area, and the area of the smaller disk is one-fourth of the original disk).

3. By the parallel axis theorem, the moment of inertia of the hole about the center of the original disk is given by:
   \[
   I_h = I_{\text{center}} + m d^2,
   \]
   where \( I_{\text{center}} \) is the moment of inertia of the smaller disk about its own center, \( m \) is the mass of the smaller disk, and \( d \) is the distance from the center of the smaller disk to the center of the original disk.

4. The moment of inertia of the smaller disk about its own center is:
   \[
   I_{\text{center}} = \frac{1}{2} \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{1}{32} MR^2.
   \]

5. The distance \( d \) from the center of the smaller disk to the center of the original disk is \( \frac{R}{2} \). Therefore, the term \( m d^2 \) is:
   \[
   m d^2 = \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)^2 = \frac{M}{4} \cdot \frac{R^2}{4} = \frac{1}{16} MR^2.
   \]

6. Adding these two contributions, we get the total moment of inertia of the hole about the center of the original disk:
   \[
   I_h = \frac{1}{32} MR^2 + \frac{1}{16} MR^2 = \frac{1}{32} MR^2 + \frac{2}{32} MR^2 = \frac{3}{32} MR^2.
   \]

7. Finally, the moment of inertia of the resulting object is:
   \[
   I = I_0 - I_h = \frac{1}{2} MR^2 - \frac{3}{32} MR^2 = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2.
   \]

Conclusion:
\[
\boxed{\frac{13}{32} MR^2}
\]