Question

A 1 kg baseball bat has a moment of inertia around a transverse axis through its center of gravity of 650 kg·cm². What is the moment of inertia of the bat about an axis through the handle of the bat if this axis is 50 cm from the center of gravity of the bat?

Answer 1

The moment of inertia of a baseball bat about an axis through the handle can be calculated using the parallel axis theorem. The moment of inertia about the handle axis is equal to the moment of inertia about the center of gravity axis plus the mass of the bat times the square of the distance between the two axes.

Step-by-step explanation:

The moment of inertia of an object depends on its mass and how the mass is distributed around the axis of rotation. In this case, the moment of inertia of the baseball bat about a handle axis can be calculated using the parallel axis theorem. The moment of inertia about the handle axis is equal to the moment of inertia about the center of gravity axis plus the mass of the object times the distance between the two axes squared.

Using the equation I(handle) = I(center) + (m * d²), where I(center) is the moment of inertia about the center of gravity axis, m is the mass of the bat, and d is the distance between the center of gravity and the handle axis, we can calculate the moment of inertia about the handle axis.

Substituting the given values, I(handle) = 650 kg·cm² + (1 kg * (50 cm)²) = 9000 kg·cm².