A uniform meter stick is pivoted to rotate about a horizontal axis through the 25-cm mark on the stick. The stick is released from rest in a horizontal position. The moment of inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by (1/12)ML². Determine the magnitude of the initial angular acceleration of the stick. A) 17 rad/s² B) 13 rad/s² C) 15 rad/s² D) 19 rad/s² E) 23 rad/s²

Question

Quizplus · Accepted Answer

The initial angular acceleration can be found using the torque and moment of inertia. The torque ($\tau$) due to gravity acts at the center of mass of the stick, which is 50 cm from the pivot for a 1-meter stick. However, since the pivot is at the 25 cm mark, the effective lever arm is 25 cm (0.25 m) from the center of mass to the pivot. The torque is $\tau = mgd$, where $m$ is the mass of the stick, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $d$ is the distance from the pivot to the center of mass (0.25 m). The moment of inertia ($I$) of the stick about the pivot is the sum of the moment of inertia about the center of mass ($\frac{1}{12}ML^2$, where $L$ is the length of the stick) plus the moment of inertia due to the parallel axis theorem ($Md^2$), where $d$ is the distance from the center of mass to the pivot. Thus, $I = \frac{1}{12}ML^2 + Md^2$. The angular acceleration ($\alpha$) can be found from $\tau = I\alpha$. Substituting the expressions for $\tau$ and $I$ and solving for $\alpha$ gives the initial angular acceleration. The correct calculation leads to the answer 17 rad/s$^2$, considering the specific distribution of mass and the forces acting on the stick.

A Uniform Meter Stick Is Pivoted to Rotate About a Horizontal