# Question about the angular speed of a rod as it falls to the ground [closed]

I was given a problem yesterday in which a bird flies into a rod and then is stunned. It gives the rod some initial angular speed (found through conservation of momentum). However, my TA then claims that you can find the the final angular speed just before it falls flat by using the torque caused by gravity. He claims that the torque caused by gravity gives the rod a constant angular acceleration. To me this does not make sense because the force of gravity would not act constantly perpendicular to the rod (lever arm) until the very last instant. A classmate I talked about this with suggested that if you integrate the equation for torque it would still work, but because it goes from pi/2 to 0 it would make the angular acceleration negative. That cannot be the case.

I'm not sure how latex works, so I will upload my work as pictures.

## closed as off-topic by John Rennie, Kyle Kanos, GiorgioP, ZeroTheHero, Jon CusterMar 17 at 2:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, GiorgioP, ZeroTheHero, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

• I was going through the textbook and I just realized that what I calculated as -mgl/2 is actually the work done by torque. That should be equal to the change in kinetic energy, which seemed promising, but ultimately leads to the final angular speed being the square root of the original angular speed minus a number. That would mean the rod slows down, which makes no sense to me. Do I need to account for transnational energy conservation as well? – dminzi Mar 15 at 18:06
• Welcome to Physics! Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – Kyle Kanos Mar 16 at 10:05

However, my TA then claims that you can find the the final angular speed just before it falls flat by using the torque caused by gravity. He claims that the torque caused by gravity gives the rod a constant angular acceleration. To me this does not make sense because the force of gravity would not act constantly perpendicular to the rod (lever arm) until the very last instant.

You are correct. Your TA is incorrect.

A classmate I talked about this with suggested that if you integrate the equation for torque it would still work, but because it goes from pi/2 to 0 it would make the angular acceleration negative. That of course, cannot be the case.

I am confused if you are trying to find the angular acceleration or the work done. Even though the angular acceleration is not constant, the equation $$\tau=I\alpha$$ is still true at all points in time. You have correctly identified the torque due to gravity about the pivot as $$\tau=\frac{mgl}{2}\cos\theta$$ so all you need to do to find the angular acceleration is apply $$\tau=I\alpha$$, which I will leave to you.

However, the integral your classmate suggests would give you the work done by the torque. The issue with the integral you have done is just not being careful with your unit conventions. If you say that $$\theta$$ increases in the counter-clockwise direction, then technically the torque due to gravity is negative. $$\tau=-\frac{mgl}{2}\cos\theta$$ This will give you the correct sign of the work being done as positive as the angle decreases.

I was going through the textbook and I just realized that what I calculated as -mgl/2 is actually the work done by torque. That should be equal to the change in kinetic energy, which seemed promising, but ultimately leads to the final angular speed being the square root of the original angular speed minus a number. That would mean the rod slows down, which makes no sense to me. Do I need to account for transnational energy conservation as well?

Now you are trying to find the final angular speed? Technically you found the work done by gravity with a sign error (as discussed above). You can easily use energy conservation to determine the final angular speed based on the work done without even thinking about the angular acceleration. Using the sign conventions as discussed above, technically you would want the angular velocity to in fact decrease. But this would mean a "become more negative" type of decrease, so that the magnitude increases. If the magnitude of your angular velocity, i.e. the speed, is becoming smaller then there is an issue. But this might be due to your mix up of sign conventions as mentioned above.

As a small piece of advice, whenever you can use energy conservation with conservative forces, do that. Trying to think about the angular acceleration is extra work for you. Of course your TA said to do it this way, but I don't have much confidence in your TA to be honest.

A good rule of thumb is that if you don't care about the time it will take for things to happen, then you should use energy conservation. This is because the energies typically only depend on position and velocity. When you need to know about the times, then you have to think about forces, torques, Newton's laws, and kinematic equations. There is more room for error, in my opinion.

• Thank you!!! After correcting the issue with the negative sign, I used W = Kf - Ki and found an answer that makes sense! – dminzi Mar 16 at 0:50
• @dminzi Please upvote all useful answers, and if one is sufficient please mark it as the accepted answer. – Aaron Stevens Mar 16 at 0:53