Lagrangian for a rod pivoting at the end of a clock hand

  • Thread starter Thread starter member 731016
  • Start date Start date
  • Tags Tags
    Eom
AI Thread Summary
The discussion revolves around deriving the equation of motion (EOM) for a rod pivoting at the end of a clock hand, specifically addressing the harmonic oscillation of ##a(t)##. The initial EOM presented is ##ml^2\ddot \phi + mRl\varphi \dot \phi\sin(\phi - \varphi t) = 0##, which is later simplified using the small angle approximation. Participants emphasize the need to linearize the equation of motion around a stable point ##\phi^*##, suggesting a Taylor expansion to retain only linear terms. The corrected EOM is identified as ##l^2 \ddot a + Rl \gamma \dot \phi a = 0##, leading to the angular frequency formula ##\omega = \sqrt{\frac{R \gamma \dot \phi}{l}}##. The conversation highlights the importance of precise notation and mathematical accuracy in deriving the correct results.
member 731016
Homework Statement
Please see belwo
Relevant Equations
##ml^2\ddot \phi + mRl\varphi \dot \phi\sin(\phi - \varphi t) = 0##
For (d),
1713422922316.png


I am confused how ##a(t)## oscillates harmonically. The EOM is ##ml^2\ddot \phi + mRl\varphi \dot \phi\sin(\phi - \varphi t) = 0##

Then using ##\phi - \varphi t = a(t)##
##l\ddot \phi + R\phi \varphi a(t) = 0##
Where I used the small angle approximation after substituting in ##a(t)##
Does anybody please know where to go from here to find ##\omega##?

Thanks!
 
Physics news on Phys.org
Simply linearise the equation of motion around ##\phi^*##. You have attempted this but need to double check your math.

Also note: It is ##\gamma##, not ##\varphi##.
 
  • Love
Likes member 731016
Orodruin said:
Simply linearise the equation of motion around ##\phi^*##. You have attempted this but need to double check your math.

Also note: It is ##\gamma##, not ##\varphi##.
Thank you for your reply @Orodruin!

##l^2\ddot \phi + Rl\phi \gamma a(t) = 0## is that please correct now?

Sorry I don't understand how you linearize the equation of motion around ##\phi^*##. I have not been taught that. Could you please explain a bit more?

Thanks!
 
ChiralSuperfields said:
Thank you for your reply @Orodruin!

##l^2\ddot \phi + Rl\phi \gamma a(t) = 0## is that please correct now?

No. Carefully consider the second term. Where did the ##\phi## come from?

ChiralSuperfields said:
Sorry I don't understand how you linearize the equation of motion around ##\phi^*##. I have not been taught that. Could you please explain a bit more?
It is essentially what you are (attempting) to do. You write ##\phi = \phi^* + a(t)## and then assume that ##a(t)## is small. Perform a Taylor expansion in ##a(t)## and keep only linear terms.
 
  • Love
Likes member 731016
Also, for the first term, you have not inserted the assumption ##\phi = \phi^* + a## ...
 
  • Love
Likes member 731016
I have solved the problem now.

EOM: ##l^2 \ddot a + Rl \gamma \dot \phi a = 0##

Angular frequency ##\omega = \sqrt{\frac{R \gamma \dot \phi}{l}}##

Thanks!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top