A point object of mass $m$ is connected to a cylinder of radius $R$ via a massless rope. At time $t = 0$ the object is moving with an initial velocity $v_0$ perpendicular to the rope, the rope has a length $L_0$, and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds $T_{max}$. Express your answers in terms of $T_{max}$, $m$, $L_0$, $R$, and $v_0$. [asy]
size(200);
real L=6;
filldraw(CR((0,0),1),gray(0.7),black);
path P=nullpath;
for(int t=0;t<370;++t)
{ 
pair X=dir(180-t)+(L-t/180)*dir(90-t);
if(X.y>L) X=(X.x,L);
P=P--X;
}
draw(P,dashed,EndArrow(size=7));
draw((-1,0)--(-1,L)--(2,L),EndArrow(size=7));
filldraw(CR((-1,L),0.25),gray(0.7),black);[/asy]What is the kinetic energy of the object at the instant that the rope breaks?

$ \textbf{(A)}\ \frac{mv_0^2}{2} $

$ \textbf{(B)}\ \frac{mv_0^2R}{2L_0} $

$ \textbf{(C)}\ \frac{mv_0^2R^2}{2L_0^2} $

$ \textbf{(D)}\ \frac{mv_0^2L_0^2}{2R^2} $

$ \textbf{(E)}\ \text{none of the above} $
1. **Initial Kinetic Energy**: At time \( t = 0 \), the object has an initial velocity \( v_0 \) perpendicular to the rope. The initial kinetic energy \( K_0 \) of the object is given by:
   \[
   K_0 = \frac{1}{2} m v_0^2
   \]

2. **Conservation of Energy**: Since the surface is frictionless and the cylinder does not rotate, there are no non-conservative forces doing work on the system. Therefore, the total mechanical energy of the system is conserved. This implies that the kinetic energy of the object remains constant throughout the motion.

3. **Tension in the Rope**: The tension in the rope provides the centripetal force necessary for the object to move in a circular path as the rope winds around the cylinder. However, since the problem states that the rope will break when the tension exceeds \( T_{max} \), we need to consider the point at which the rope breaks.

4. **Kinetic Energy at the Breaking Point**: Since no work is done on the object (as there is no friction and the cylinder does not rotate), the kinetic energy of the object remains the same as its initial kinetic energy. Therefore, the kinetic energy of the object at the instant the rope breaks is:
   \[
   K = \frac{1}{2} m v_0^2
   \]

Conclusion:
\[
\boxed{\frac{mv_0^2}{2}}
\]