91Ó°ÊÓ

The moment of inertia is like the rotational equivalent of mass for linear motion. It measures how difficult it is to change the rotational motion of an object. For a thin rod of length \( L \) and mass \( m \), the moment of inertia about an axis through one end is given by \[ I = \frac{1}{3} m L^2 \].
In this exercise, the rod has a length of \( 0.75 \, \text{m} \) and a mass of \( 0.42 \, \text{kg} \). By substituting these values into the formula, we calculate the moment of inertia:

• \( I = \frac{1}{3} (0.42 \, \text{kg})(0.75 \, \text{m})^2 = 0.07875 \, \text{kg} \, \text{m}^2 \).

This value tells us how much resistance the rod has to being swung around the pivot at one end. The greater the moment of inertia, the more effort it takes to rotate the object.

Kinetic energy is the energy an object possesses due to its motion. For rotational motion, kinetic energy is determined by both the moment of inertia and the angular speed. The formula to calculate the kinetic energy for a rotating object is \[ KE = \frac{1}{2} I \omega^2 \], where \( \omega \) is the angular speed.
The rod swings with an angular speed of \( 4.0 \, \text{rad/s} \), and using the earlier calculated moment of inertia \( I = 0.07875 \, \text{kg} \, \text{m}^2 \), the kinetic energy at its lowest position can be computed:

• \( KE = \frac{1}{2} (0.07875 \, \text{kg} \, \text{m}^2)(4.0 \, \text{rad/s})^2 = 0.63 \, \text{J} \).

This kinetic energy indicates how much energy the rod has due to its rotational motion at the point of its maximum speed, which is at the lowest point in its swing.

The concept of conservation of energy states that energy in a closed system must remain constant. In the context of this pendulum-like motion, the rod's kinetic energy at the lowest point transforms to potential energy as it rises. The formula relating kinetic energy to potential energy is given by \[ KE = PE \] or \[ \frac{1}{2} I \omega^2 = mgh \]. Here, \( h \) represents the height the rod's center of mass is elevated.
To determine how high the rod's center of mass rises, we rearrange the equation to solve for \( h \):

• \( h = \frac{(0.63 \, \text{J})}{(0.42 \, \text{kg} \cdot 9.8 \, \text{m/s}^2)} = 0.153 \, \text{m} \).

This calculation highlights that the energy initially as kinetic at the lowest point is gradually converted into potential energy as the rod ascends, lifting its center of mass by 0.153 meters.