A uniform rod (length = 2.0 m) is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through the rod at a point 0.50 m from one end of the rod. If the rod is released from rest in a horizontal position, what is the angular speed of the rod as it rotates through its lowest position?
A) 3.5 rad/s
B) 3.8 rad/s
C) 4.1 rad/s
D) 2.0 rad/s
E) 5.6 rad/s

Question

Quizplus · Accepted Answer

The angular speed of the rod as it rotates through its lowest position can be found using the principle of conservation of energy. Initially, the rod has gravitational potential energy when it is horizontal, and this energy is converted into rotational kinetic energy as it rotates to the vertical position. The potential energy (PE) at the start is given by $PE = mgh$, where $m$ is the mass of the rod, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the height of the center of mass from the pivot point. The center of mass of the rod is at its midpoint, so for a 2.0 m rod with the pivot 0.5 m from one end, the center of mass is 1.0 m from the pivot. Thus, $h = 1.0$ m.The rotational kinetic energy (RKE) at the lowest position is given by $RKE = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the rod about the pivot point, and $\omega$ is the angular speed.The moment of inertia of a rod about an axis through its end is given by $I = \frac{1}{3}mL^2$, but since the axis is not at the end, we use the parallel axis theorem to adjust this: $I = \frac{1}{3}mL^2 + md^2$, where $L = 2.0$ m is the length of the rod, and $d = 0.5$ m is the distance from the pivot to the rod's end. However, this formula assumes the pivot is at the end. For our case, the correct moment of inertia considering the pivot location is $I = \frac{1}{12}m(2.0)^2 + m(0.5)^2$, because the general formula for a rod about its center is $\frac{1}{12}mL^2$ and we adjust for the pivot being 0.5 m from the end.Setting the initial potential energy equal to the final rotational kinetic energy, $mgh = \frac{1}{2}I\omega^2$, and solving for $\omega$, we find that the mass $m$ cancels out. Plugging in the values for $g$, $h$, and $I$, and solving for $\omega$ gives us the angular speed.After calculating with $g = 9.8$ m/s$^2$, $h = 1.0$ m, and the moment of inertia $I$ calculated as described, you find that $\omega$ is approximately 4.1 rad/s, which corresponds to choice C.

A Uniform Rod (Length = 2