For any given triangle $A_0B_0C_0$ consider a sequence of triangles constructed as follows: 
a new triangle $A_1B_1C_1$ (if any) has its sides (in cm) that equal to the angles of $A_0B_0C_0$ (in radians). Then for $\vartriangle  A_1B_1C_1$  consider a new triangle $A_2B_2C_2$ (if any) constructed in the similar พay, i.e., $\vartriangle A_2B_2C_2$ has its sides (in cm) that equal to the angles of $A_1B_1C_1$ (in radians), and so on.
Determine for which initial triangles $A_0B_0C_0$ the sequence never terminates.
To determine for which initial triangles \( \triangle A_0B_0C_0 \) the sequence never terminates, we need to analyze the transformation process and the conditions under which it can continue indefinitely.

1. **Initial Triangle \( \triangle A_0B_0C_0 \)**:
   - Let the angles of \( \triangle A_0B_0C_0 \) be \( \alpha_0, \beta_0, \gamma_0 \) in radians.
   - The sides of \( \triangle A_1B_1C_1 \) will be \( \alpha_0, \beta_0, \gamma_0 \) in centimeters.

2. **Constructing \( \triangle A_1B_1C_1 \)**:
   - For \( \triangle A_1B_1C_1 \) to be a valid triangle, the sum of any two sides must be greater than the third side. This is the triangle inequality theorem.
   - Therefore, we need:
     \[
     \alpha_0 + \beta_0 > \gamma_0, \quad \beta_0 + \gamma_0 > \alpha_0, \quad \gamma_0 + \alpha_0 > \beta_0
     \]
   - Since \( \alpha_0, \beta_0, \gamma_0 \) are the angles of a triangle, they satisfy \( \alpha_0 + \beta_0 + \gamma_0 = \pi \). This ensures that the above inequalities are satisfied.

3. **Constructing \( \triangle A_2B_2C_2 \)**:
   - The angles of \( \triangle A_1B_1C_1 \) will be determined by the sides \( \alpha_0, \beta_0, \gamma_0 \).
   - Let the angles of \( \triangle A_1B_1C_1 \) be \( \alpha_1, \beta_1, \gamma_1 \).
   - The sides of \( \triangle A_2B_2C_2 \) will be \( \alpha_1, \beta_1, \gamma_1 \) in centimeters.

4. **Continuing the Sequence**:
   - For the sequence to never terminate, each subsequent triangle must be a valid triangle.
   - This means that the angles of each triangle must always be such that the sides of the next triangle satisfy the triangle inequality.

5. **Special Case: Isosceles Triangle**:
   - Consider an isosceles triangle with sides \( L, L, \ell \) where \( \ell \ll L \).
   - The angles of this triangle will be approximately \( \alpha \approx \frac{\pi - \theta}{2} \) and \( \beta = \gamma \approx \theta \) where \( \theta \) is very small.
   - The sides of the next triangle will be approximately \( \frac{\pi - \theta}{2}, \theta, \theta \).
   - As \( \theta \) is very small, the new triangle will still satisfy the triangle inequality, and the process can continue.

6. **General Condition**:
   - For the sequence to never terminate, the initial triangle \( \triangle A_0B_0C_0 \) must be such that the angles of each subsequent triangle always form a valid triangle.
   - This is generally true for any non-degenerate triangle, as the angles will always sum to \( \pi \) and satisfy the triangle inequality.

Conclusion:
The sequence of triangles will never terminate for any non-degenerate initial triangle \( \triangle A_0B_0C_0 \).

The final answer is any non-degenerate initial triangle \( \boxed{ \triangle A_0B_0C_0 } \).