A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML²/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released? A) 9.8 rad/s² B) 7.4 rad/s² C) 8.4 rad/s² D) 5.9 rad/s² E) 6.5 rad/s²

Question

Quizplus · Accepted Answer

The angular acceleration can be found using the equation for torque ($\tau = I\alpha$) and knowing that the torque due to gravity at the instant of release is $mg\frac{L}{2}\cos(\theta)$, where $L$ is the length of the rod, $m$ is its mass, $\theta$ is the angle with the horizontal, and $g$ is the acceleration due to gravity. The moment of inertia ($I$) of a rod about an end is $\frac{1}{3}mL^2$. Substituting these into the equation for torque gives $\alpha = \frac{3g\cos(\theta)}{2L}$. At $\theta = 37^\circ$, $\cos(37^\circ) \approx 0.8$, and substituting $g = 9.8 \, \text{m/s}^2$ and $L = 1.2 \, \text{m}$, we find $\alpha \approx 9.8 \, \text{rad/s}^2$, indicating that the angular acceleration of the rod at the moment it is released is approximately 9.8 rad/s².

A Thin Uniform Rod (Length = 1