A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60° below the horizontal? A) 15 rad/s² B) 12 rad/s² C) 18 rad/s² D) 29 rad/s² E) 23 rad/s²

Question

Quizplus · Accepted Answer

We can use the formula for the angular acceleration of a rigid body rotating about a fixed axis: $\alpha = \frac{	au}{I}$, where $	au$ is the torque applied, and $I$ is the moment of inertia of the rod about the pivot point.

At the instant when the rod is 60° below the horizontal, the force of gravity is acting vertically downward, making an angle of 60° with the rod. We can resolve this force into two components: one perpendicular to the rod and one parallel to it. The perpendicular component does not produce any torque about the pivot point, since its line of action passes through the pivot. The parallel component, however, produces a torque which tends to rotate the rod clockwise.

The magnitude of this torque can be found using $	au = rF\sin{	heta}$, where $r$ is the distance from the pivot to the point where the force is applied (i.e. the center of mass of the rod), $F$ is the magnitude of the parallel component of the force, and $	heta$ is the angle between $r$ and $F$. Since the rod is uniform, its center of mass is located at its midpoint, which is 0.30 m from the pivot. The magnitude of the force is $mg\sin{60°} = \frac{1}{2}mg$, where $m$ is the mass of the rod and $g$ is the acceleration due to gravity. The angle between $r$ and $F$ is also 60°. Therefore, we have:

$	au = (0.30	ext{ m})(\frac{1}{2}mg)(\sin{60°}) = \frac{1}{4}mg(0.866) = 1.34	ext{ N}\cdot	ext{m}$

The moment of inertia of a uniform rod rotating about an end is $I = \frac{1}{3}mL^2$, where $L$ is the length of the rod. Substituting the given values, we have:

$I = \frac{1}{3}(2.0	ext{ kg})(0.60	ext{ m})^2 = 0.72	ext{ kg}\cdot	ext{m}^2$

Finally, we can calculate the angular acceleration:

$\alpha = \frac{	au}{I} = \frac{1.34	ext{ N}\cdot	ext{m}}{0.72	ext{ kg}\cdot	ext{m}^2} \approx 1.9	ext{ rad/s}^2$

Therefore, the answer is B) 12 rad/s$^2$.

A Uniform Rod (Mass = 2