Pivoting rod analyzed about a rotating axis

Question

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?

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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 $v=\sqrt{3gL}$

Now for the axis of rotation as the center of mass:

I reasoned $v_{AOR}$ ($v_{cm}$) to be v/2, as both the free end and the center of mass have the same angular velocities, yet the axis of rotation is half the distance from the pivot point as the free end.

The moment of inertia is $\frac{1}{12}mL^2$

I believe $\omega = \frac{v}{L/2}$. Initial intuition is that the free end has velocity v by definition (at least in the inertial reference frame), and the distance between the axis of rotation and the free end is L/2 so by $v=r\omega$ (and of course the vectors being orthogonal) $\omega = \frac{v}{L/2}$. I got the same result when parametrizing both the free end and the axis of rotation as $\overrightarrow{r_2}=\left\langle Lcos(\omega t), -Lsin(\omega t) \right\rangle$ and $\overrightarrow{r_1}=\left\langle L/2cos(\omega t), -L/2sin(\omega t) \right\rangle$, taking the difference of these two as $\overrightarrow{r}$, taking the derivative of $\overrightarrow{r}$ and then finding the magnitude of that quantity.

For $w_{AOR}$ I got $\frac{1}{8}mgL$. For brevity I will just post summary steps, but feel free to ask for more work. I found the centripetal acceleration of the center of mass as a function of $\theta$ to be $\frac{3}{2}gsin(\theta)$ and the tangential acceleration to be $\frac{3}{4}gcos(\theta)$. Converting to Cartesian coordinates $$\overrightarrow{a} = \frac{3}{2}gsin(\theta)\left\langle -cos(\theta), sin(\theta), 0 \right\rangle + \frac{3}{4}gcos(\theta)\left\langle -sin(\theta), -cos(\theta), 0 \right\rangle$$ Then, by Newton's second law $$\left\langle 0,-mg,0 \right\rangle + \overrightarrow{F} = m\overrightarrow{a}$$ where $\overrightarrow{F}$ is the force exerted by the pin. After solving the equation above for the force, I find the torque produced by this force on the axis of rotation by taking the cross product between the moment arm and this pin force: $$\left\langle -\frac{L}{2}cos(\theta), \frac{L}{2}sin(\theta), 0 \right\rangle \times \left\langle \frac{-9mgsin(\theta)cos(\theta)}{4}, \frac{mg(9sin^2(\theta)+1)}{4}, 0 \right\rangle$$$$=\left\langle 0,0, \frac{-mgLcos(\theta)}{8} \right\rangle$$ Taking the integral from $\theta =0$ to $\theta = \pi/2$ of this torque $d\overrightarrow{\theta}$ yields $w_= \frac{1}{8}mgL$

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 $v=\sqrt{3gL}$.

Answer 1

Two mistakes were made. First, the translational work done by the pin was neglected. By performing similar calculations as was done for the rotational work by the pin, the translation work is found to be $-\frac{1}{8}mgL$. Second, the angular velocity of the free end about the axis of rotation is $v/ L$. By considering both of these mistakes, the conservation of energy equation will produce $v=\sqrt{3gL}$.

The reason that the angular velocity of the free end about the axis of rotation is $v/L$ is that in the fixed center of mass frame, the velocity $v'$ of the free end is indeed $\frac{L}{2}\omega$. However, to convert to the inertial reference frame, you have to add back the velocity of the center of mass: $v' + v_{cm} = v$ and $v_{cm}=v/2$, so $v'=v/2$ and thus, $\frac{v}{2}=\frac{L}{2}\omega$.

Answer 2

So, I will show you the solution using the concept of Conservation of Energy.

So, When the rod is released, it's centre of mass travels a vertical distance of "h" which leads to the loss in Gravitational Potential Energy equal to

ΔU = Mgh (Here "h" will be equal to L/2 where L is the length of the rod)

This lost potential energy will be converted to Kinetic Energy, but it is difficult to write the kinetic energy of the rod because every part of the rod is moving with different velocities.

But we know that the rod is performing pure rotational motion about the pivot, therefore we can write the kinetic energy of rod as rotational kinetic energy about pivot, so the equation of rotational kinetic energy comes out to be

                     KE(Kinetic Energy)= 1/2*I*ω^2


                    ΔKE(change in kinetic energy)=KE(final)- KE(initial)

Now, ΔU = ΔKE

                         Mgh=1/2 * M*L^2*1/3(MOI) * ω^2 ---{1}

MOI- Moment of inertia(here of rod about one of it's end)

On solving equation {1} we get the value of ω as

                            ω=[3g/L]^0.5

Now, To get the speed of the free end of the Rod multiply ω with distance of free end from the point of suspension(here = L)

Therefore we finally get V=[3gL]^0.5