Problem 1359

1. To determine the angle $\theta_g$ at which the tension in the rod is greatest, we need to analyze the forces acting on the point mass $m$ during its motion. The forces acting on the mass are:
- The gravitational force $mg$ acting downward.
- The tension $T$ in the rod acting along the rod towards the pivot.

2. The tension in the rod can be expressed as the sum of the centripetal force required to keep the mass moving in a circular path and the component of the gravitational force acting along the rod. The centripetal force is given by $\frac{mv^2}{L}$, where $v$ is the tangential velocity of the mass, and the component of the gravitational force along the rod is $mg \cos \theta$.

3. Therefore, the tension $T$ in the rod at any angle $\theta$ can be written as:
$$T = \frac{mv^2}{L} + mg \cos \theta$$

4. To find the angle $\theta_g$ at which the tension is greatest, we need to consider the behavior of the tangential velocity $v$ as a function of $\theta$. The energy conservation principle can be used here. The total mechanical energy (potential + kinetic) of the pendulum is conserved. At the highest point (release point) $\theta = \theta_0$, the pendulum has maximum potential energy and zero kinetic energy. At the lowest point $\theta = 0$, the pendulum has maximum kinetic energy and minimum potential energy.

5. The potential energy at an angle $\theta$ is given by:
$$U = mgh = mgL(1 - \cos \theta)$$
where $h = L(1 - \cos \theta)$ is the height of the mass above the lowest point.

6. The kinetic energy at an angle $\theta$ is given by:
$$K = \frac{1}{2}mv^2$$

7. Using conservation of energy, the total energy at the release point $\theta_0$ is equal to the total energy at any other point $\theta$:
$$mgL(1 - \cos \theta_0) = \frac{1}{2}mv^2 + mgL(1 - \cos \theta)$$

8. Solving for $v^2$, we get:
$$v^2 = 2gL(\cos \theta - \cos \theta_0)$$

9. Substituting $v^2$ into the expression for tension $T$, we get:
$$T = \frac{m \cdot 2gL(\cos \theta - \cos \theta_0)}{L} + mg \cos \theta$$
$$T = 2mg(\cos \theta - \cos \theta_0) + mg \cos \theta$$
$$T = mg(2\cos \theta - 2\cos \theta_0 + \cos \theta)$$
$$T = mg(3\cos \theta - 2\cos \theta_0)$$

10. To maximize the tension $T$, we need to maximize the term $3\cos \theta - 2\cos \theta_0$. Since $\cos \theta$ is maximum when $\theta = 0$, the tension $T$ is maximized at $\theta = 0$.

Conclusion:
The tension in the rod is greatest at $\theta_g = 0$.

The final answer is $\boxed{\textbf{(B) } \theta_g = 0}$