# Angular momentum or work-energy balance?

1. Aug 11, 2013

### cambo86

[Broken]
The apparatus above has an initial angular velocity of $\omega_{1}$ as the rods are released. I need to find the angular velocity $\omega_{2}$ of the apparatus at the bottom.

I've tried 3 methods.
First I tried a work-energy balance where I included a gravitational potential energy.
$KE_{1} + PE = KE_{2}$
$0.5 I_{1}\omega_{1}^{2} + mgh = 0.5 I_{2}\omega_{2}^{2}$

Then I figured that the potential energy is going to only contribute to an increase in velocity of the arms falling down so I dropped the PE term and got,
$0.5 I_{1}\omega_{1}^{2} = 0.5 I_{2}\omega_{2}^{2}$

I'm just wondering which is correct?

Then I spoke to another student in the class and he said he used conservation of angular momentum. So, he got,
$I_{1}\omega_{1} = I_{2}\omega_{2}$

Last edited by a moderator: May 6, 2017
2. Aug 11, 2013

### cambo86

The angular momentum I have to solve for is indicated by N in the picture.

3. Aug 11, 2013

### TSny

Hello.

There are no external torques acting on the apparatus. So, total angular momentum about the vertical rotation axis will be conserved. So, $I_1\omega_1 = I_2\omega_2$ will get the answer.

Energy conservation is interesting here. If you assume no dissipation of mechanical energy due to friction in the horizontal pins through the ends of the rods, then total mechanical energy will be conserved. When the rods reach the horizontal position, they will be rotating about the pins as well as rotating about the vertical axis. So, you would have to write the equation of conservation of energy as $$KE_1 + PE = KE_2 + KE'$$ where $KE_1$ and $KE_2$ are the initial and final KE's associated with rotation about the vertical axis and $KE'$ is the kinetic energy associated with the rotation of the rods about the pins in the ends of the rods at the instant the rods are horizontal. Both $KE_2$ and $KE'$ are unknowns, so the energy equation alone does not allow you to get the answer.

Although you might expect $KE'$ to equal $PE$, it turns out that $KE'$ is greater than $PE$. So, when the arms reach the horizontal position, they have more KE of rotation about the horizontal pins than the amount of initial PE. Thus, it must be that some of the initial $KE_1$ about the vertical axis is transferred to $KE'$. This in turn means that $KE_2 < KE_1$; i.e., the final KE about the vertical axis is less than the initial KE about the vertical axis.

You can verify this by first using conservation of angular momentum to get $\omega_2$ and then calculating $KE_2$ to see that $KE_2 < KE_1$.