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When solving a classic rotational dynamics problem — a uniform rod of length L and mass m, pivoted at one end by frictionless pin and released from rest in the horizontal position rotates under the force of gravity — I attempted to calculate the speed of the free end once the rod reaches the vertical position using conservation of energy with two different axes of rotation:

• About the pivot point: Using standard calculations, the speed of the free end when the rod is vertical is $v=\sqrt{3gL}$
• About the center of mass: I redefined the axis of rotation to pass through the center of mass, where the pin at one end exerts a nonzero torque. Despite detailed calculations, my result for v does not match the expected $v=\sqrt{3gL}$, but instead yields $v=\sqrt{\frac{15}{7}gL}$

When solving a classic rotational dynamics problem — a uniform rod of length $L$ and mass $m$, pivoted at one end by frictionless pin and released from rest in the horizontal position rotates under the force of gravity — I attempted to calculate the speed of the free end once the rod reaches the vertical position using conservation of energy with two different axes of rotation:

• About the pivot point: Using standard calculations, the speed of the free end when the rod is vertical is $v=\sqrt{3gL}$

• About the center of mass: I redefined the axis of rotation to pass through the center of mass, where the pin at one end exerts a nonzero torque. Despite detailed calculations, my result for $v$ does not match the expected $v=\sqrt{3gL}$, but instead yields $v=\sqrt{\frac{15}{7}gL}$

A uniform rod of length L and mass m is pivoted about a frictionless pin through one end such that the rod rotates due to gravity from rest in a horizontal position through a vertical position. Let $\theta$ be the angle of the rod from the horizontal such that $\theta=0$ corresponds to the rod being horizontal and $\theta=\pi/2$ to the rod being vertical. If I let the axis of rotation be the pivot point, then straightforward calculations of conservation of energy reveal that at the vertical position the speed of the free end of the rod is $v=\sqrt{3gL}$. However, as a challenge I sought to reproduce this result using the center of mass of the rod as the axis of rotation, where now the pin (still at one end) would do work as the torque is nonzero about the center of mass. First, my interpretation of conversation of energy for a generalized axis of rotation:
$$\frac{1}{2}mv^2_{AOR} + \frac{1}{2}I\omega^2=mgh + w_{AOR}$$ where $v_{AOR}$ is the velocity of the axis of rotation, $\omega$ is the angular velocity of the rod emphasized textabout the axis of rotation, h is the change in height of the center of mass, and w is the work done to the axis of rotation by the pin. Note that if the axis of rotation is taken to be the pivot point, then $v_{AOR}$=0, $I=\frac{1}{3}mL^2$, $\omega =\frac{v}{L}$ and $w_{AOR}=0$, and solving for v yields $v=\sqrt{3gL}$

However, when putting this all together: $$\frac{1}{2}m(v/2)^2 + \frac{1}{2}(\frac{1}{12}mL^2)(\frac{v}{L/2})^2 = \frac{1}{2}mgL + \frac{1}{8}mgL$$ and solving for the velocity of the free end at the vertical position yields $v=\sqrt{\frac{15}{7}gl}$, not the correct answer of $v=\sqrt{3gL}$. Please help me address where I went wrong

When solving a classic rotational dynamics problem — a uniform rod of length L and mass m, pivoted at one end by frictionless pin and released from rest in the horizontal position rotates under the force of gravity — I attempted to calculate the speed of the free end once the rod reaches the vertical position using conservation of energy with two different axes of rotation:

• About the pivot point: Using standard calculations, the speed of the free end when the rod is vertical is $v=\sqrt{3gL}$
• About the center of mass: I redefined the axis of rotation to pass through the center of mass, where the pin at one end exerts a nonzero torque. Despite detailed calculations, my result for v does not match the expected $v=\sqrt{3gL}$, but instead yields $v=\sqrt{\frac{15}{7}gL}$

Here is where I believe my reasoning might have gone astray:

• Interpreting work done by the pin: I derived $w=\frac{1}{8}mgL$ by integrating the torque exerted by the pin about the center of mass. Is this approach correct for determining the work in this scenario? Could there be a fundamental misunderstanding in how work is treated when the axis of rotation itself is in motion?
• Defining angular velocity and linear velocity relationships: I assumed $\omega = \frac{v}{L/2}$ for the angular velocity of free end with respect to the center of mass (axis of rotation) as v is the velocity of the free end and the distance to the center of mass is L/2. Is this a valid application of v=rω, or does the moving axis of rotation introduce additional considerations as a non-inertial reference frame?

To make this more general, I am seeking clarification on how to correctly apply conservation of energy and account for work when using a non-fixed axis of rotation, particularly in cases where the axis is moving due to external forces. Could someone explain the proper methodology and common pitfalls in such scenarios?

Explanation of my work:

First, my interpretation of conversation of energy for a generalized axis of rotation:
$$\frac{1}{2}mv^2_{AOR} + \frac{1}{2}I\omega^2=mgh + w_{AOR}$$ where $v_{AOR}$ is the velocity of the axis of rotation, $\omega$ is the angular velocity of the rod about the axis of rotation, h is the change in height of the center of mass, and w is the work done to the axis of rotation by the pin. Note that if the axis of rotation is taken to be the pivot point, then $v_{AOR}$=0, $I=\frac{1}{3}mL^2$, $\omega =\frac{v}{L}$ and $w_{AOR}=0$, and solving for v yields $v=\sqrt{3gL}$

However, when putting this all together: $$\frac{1}{2}m(v/2)^2 + \frac{1}{2}(\frac{1}{12}mL^2)(\frac{v}{L/2})^2 = \frac{1}{2}mgL + \frac{1}{8}mgL$$ and solving for the velocity of the free end at the vertical position yields $v=\sqrt{\frac{15}{7}gL}$, not the correct answer of $v=\sqrt{3gL}$.