A nonuniform 2.0-kg rod is 2.0 m long.The rod is mounted to rotate freely about a horizontal axis perpendicular to the rod that passes through one end of the rod.The moment of inertia of the rod about this axis is 4.0 kg.m².The center of mass of the rod is 1.2 m from the axis.If the rod is released from rest in the horizontal position,what is its angular speed as it swings through the vertical position?

Question

Quizplus · Accepted Answer

The correct answer is A because the angular speed of the rod as it swings through the vertical position is 3.4 rad/s. This can be calculated using the conservation of energy, which states that the total energy of the system is constant. In this case, the total energy of the system is the sum of the kinetic energy of the rod and the potential energy of the rod due to gravity. When the rod is released from rest in the horizontal position, it has no kinetic energy and all of its energy is potential energy. As the rod swings down, it gains kinetic energy and loses potential energy. At the vertical position, the rod has maximum kinetic energy and zero potential energy. Therefore, the angular speed of the rod at the vertical position can be calculated using the following equation:```1/2 * I * w^2 = m * g * h```where:* I is the moment of inertia of the rod* w is the angular speed of the rod* m is the mass of the rod* g is the acceleration due to gravity* h is the height of the rod above the vertical positionPlugging in the given values, we get:```1/2 * 4.0 kgF1F1F1S1F1F1F10m² * w^2 = 2.0 kg * 9.8 m/s^2 * 1.2 m```Solving for w, we get:```w = 3.4 rad/s```

A Nonuniform 2