A uniform rod of length (L = 2.0 m) and mass (M = 1.5 kg) is pivoted about a horizontal frictionless pin through one end. The rod is released from rest at an angle of 30° below the horizontal. What is the angular speed of the rod when it passes through the vertical position? (The moment of inertia of the rod about the pin is 2.0 kg⋅m².) A) 3.5 rad/s B) 2.7 rad/s C) 3.1 rad/s D) 2.3 rad/s E) 1.6 rad/s

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Quizplus · Accepted Answer

The angular speed of the rod when it passes through the vertical position can be found using the principle of conservation of energy. Initially, the rod has potential energy and no kinetic energy. When it reaches the vertical position, all the potential energy has been converted into rotational kinetic energy. The initial potential energy (PE_initial) is given by $PE = Mgh$, where $h$ is the height of the center of mass from the lowest position. For a rod pivoted at one end, the center of mass is at $L/2$, so $h = L/2 \sin(30°) = L/4$. The rotational kinetic energy (KE_rotational) when the rod is vertical is given by $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular speed.Setting the initial potential energy equal to the final kinetic energy gives:$$MgL/4 = \frac{1}{2}I\omega^2$$Given $M = 1.5\,kg$, $L = 2.0\,m$, $g = 9.8\,m/s^2$, and $I = 2.0\,kg\cdot m^2$, we can solve for $\omega$:$$1.5 \times 9.8 \times 2.0/4 = \frac{1}{2} \times 2.0 \times \omega^2$$$$7.35 = \omega^2$$$\omega = \sqrt{7.35} \approx 2.7\,rad/s$Therefore, the angular speed of the rod when it passes through the vertical position is approximately 2.7 rad/s.

A Uniform Rod of Length (L = 2