A uniform rod (mass = 1.5 kg) is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a horizontal position. What is the angular speed of the rod when the rod makes an angle of 30° with the horizontal? (The moment of inertia of the rod about the pin is 2.0 kg⋅m²). A) 2.2 rad/s B) 3.6 rad/s C) 2.7 rad/s D) 3.1 rad/s E) 1.8 rad/s

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Quizplus · Accepted Answer

The angular speed of the rod can be found using the principle of conservation of energy. Initially, the rod has potential energy and no kinetic energy. When it reaches an angle of 30°, it has both kinetic and potential energy. The initial potential energy (PE_initial) when the rod is horizontal is given by $PE = mgh$, where $h$ is the height of the center of mass from the pivot. Since the rod is uniform and pivoted at one end, the center of mass is at the midpoint, so $h = L/2 = 1.0 \, \text{m}$ initially. When the rod is at 30°, the height of the center of mass ($h'$) is $L/2 \cos(30°)$. The change in potential energy ($\Delta PE$) is $mgh - mg(L/2 \cos(30°))$. This change is equal to the change in kinetic energy, which is rotational in this case, given by $1/2 I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular speed.Given:- $m = 1.5 \, \text{kg}$- $L = 2.0 \, \text{m}$- $I = 2.0 \, \text{kg} \cdot \text{m}^2$- $g = 9.8 \, \text{m/s}^2$Solving for $\omega$ using the conservation of energy principle and substituting the given values, we find that the angular speed $\omega$ corresponds to the value closest to 2.7 rad/s, which is option C.

A Uniform Rod (Mass = 1