An open jar of water moves in a vertical circle of radius 0.50 m with a frequency that is small enough to put the water on the verge of falling out of the jar at the top of the circle. If the same demonstration is only `3.7m//s^(2)`, what is the change in the circling frequency to again put the water on the verge of falling out at the top point?

Explanation

To solve the problem, we need to determine the change in the angular frequency (ω) of the jar of water when moving from Earth to Mars, where the gravitational acceleration is different.

### Step-by-Step Solution:

1. **Understand the Forces at the Top of the Circle**:
At the top of the circular path, the water is on the verge of falling out. The centripetal force required to keep the water in circular motion is provided by the gravitational force acting on it. Thus, we can set up the equation:
\[
\frac{m \cdot \omega^2 \cdot r}{g} = m \cdot g
...