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]
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for(int t=0;t<370;++t)
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pair X=dir(180-t)+(L-t/180)*dir(90-t);
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filldraw(CR((-1,L),0.25),gray(0.7),black);[/asy]What is the length (not yet wound) of the rope?

$ \textbf{(A)}\ L_0 - \pi R $

$ \textbf{(B)}\ L_0 - 2 \pi R$

$ \textbf{(C)}\ L_0 - \sqrt{18} \pi R $

$ \textbf{(D)}\ \frac{mv_0^2}{T_{max}} $

$ \textbf{(E)}\ \text{none of the above} $
1. **Identify the forces and motion involved:**
   - The object of mass \( m \) is moving in a circular path around the cylinder.
   - The tension \( T \) in the rope provides the centripetal force required for circular motion.
   - The centripetal force is given by \( T = \frac{mv^2}{r} \), where \( r \) is the radius of the circular path and \( v \) is the velocity of the object.

2. **Determine the relationship between the tension and the radius:**
   - Initially, the length of the rope is \( L_0 \) and the radius of the cylinder is \( R \).
   - As the rope winds around the cylinder, the effective length of the rope decreases.
   - The radius \( r \) of the circular path at any time is given by \( r = L_0 - l \), where \( l \) is the length of the rope wound around the cylinder.

3. **Find the condition when the rope breaks:**
   - The rope will break when the tension \( T \) exceeds \( T_{max} \).
   - At the breaking point, \( T = T_{max} \).

4. **Express the velocity in terms of the given parameters:**
   - Using the centripetal force equation at the breaking point:
     \[
     T_{max} = \frac{mv^2}{r}
     \]
   - Solving for \( v \):
     \[
     v = \sqrt{\frac{T_{max} r}{m}}
     \]

5. **Determine the length of the rope not yet wound:**
   - The initial kinetic energy of the object is \( \frac{1}{2}mv_0^2 \).
   - As the object moves, the kinetic energy is converted into the work done against the tension in the rope.
   - The work done by the tension is \( T_{max} \times (L_0 - r) \).
   - Equating the initial kinetic energy to the work done:
     \[
     \frac{1}{2}mv_0^2 = T_{max} (L_0 - r)
     \]
   - Solving for \( L_0 - r \):
     \[
     L_0 - r = \frac{mv_0^2}{2T_{max}}
     \]

6. **Find the length of the rope not yet wound:**
   - The length of the rope not yet wound is \( L_0 - r \).
   - Using the expression derived:
     \[
     L_0 - r = \frac{mv_0^2}{2T_{max}}
     \]

The final answer is \(\boxed{\frac{mv_0^2}{2T_{max}}}\)