The given problem involves the oscillation of a physical pendulum. The body has a mass, moment of inertia, and distance from the center of mass to the pivot point. We're asked to calculate the period of oscillation.

Given:

• Mass of the body $m = 170 \, \text{g} = 0.170 \, \text{kg}$
• Moment of inertia $I = 0.00614 \, \text{kg} \cdot \text{m}^2$
• Distance from center of mass to the pivot $d = 19.0 \, \text{cm} = 0.190 \, \text{m}$
• Gravitational acceleration $g = 9.81 \, \text{m/s}^2$

For a physical pendulum, the period $T$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{m g d}}$

Substituting the given values:

$T = 2\pi \sqrt{\frac{0.00614}{0.170 \times 9.81 \times 0.190}}$

Now, let's calculate that step-by-step:

$T = 2\pi \sqrt{\frac{0.00614}{0.3161373}}$ $T = 2\pi \sqrt{0.019424}$ $T = 2\pi \times 0.13937$ $T = 0.8754 \, \text{seconds}$

Final Answer:

The period of oscillation of the body is approximately 0.875 seconds.

Would you like further clarification on this? Here are some related questions:

1. How does increasing the mass affect the period of oscillation?
2. What is the difference between a simple pendulum and a physical pendulum?
3. How would the period change if the center of mass were closer to the pivot point?
4. Why do we use the small angle approximation in this case?
5. What happens to the period if we change the moment of inertia?

Tip: The moment of inertia is a crucial factor in rotational systems. It determines how the mass is distributed relative to the pivot point.