Question

The following equation of motion describes the rolling motion of a boat:

IGθ + mgh sinθ = 0
Measurements of a certain boat's roll period give a value of 5 s. The boat's metacentric height is h=0.3 m. How much change in the metacentric height must be made for the roll period to be increased to 10 s?

Answer 1

To increase the roll period from 5s to 10s, the metacentric height must be quadrupled since the period is proportional to the square root of the metacentric height.

Step-by-step explanation:

The question involves the physics of oscillatory motion, specifically the rolling motion of a boat as described by its equation of motion. To increase the roll period from 5 seconds to 10 seconds, we need to understand the relationship between the roll period (T) and the metacentric height (h). The roll period of a boat is given by the formula T = 2π√(I/(mgh)), where I is the moment of inertia, m is the mass of the boat, g is the acceleration due to gravity, and h is the metacentric height.

Since the question asks for the change in metacentric height to achieve a specific roll period, we can rearrange this equation to solve for h. Given the roll period is proportional to the square root of the metacentric height, doubling the period (from 5 s to 10 s) would require quadrupling the value inside the square root, meaning the new metacentric height needs to be four times the initial value.