Kinematics problem: sliding bar.

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SUMMARY

The discussion focuses on solving a kinematics problem involving a sliding bar of length l, where point A moves at a constant velocity to the right. The goal is to determine the angle theta as a function of the position xA. The conversation highlights the use of related rates to find the angular velocity and angular acceleration of the bar about point A, given the derivative dx/dt. An error in the rearrangement of the cosine equation is also noted, emphasizing the complexity of the algebra and trigonometry involved.

PREREQUISITES
  • Understanding of kinematics principles
  • Familiarity with angular motion concepts
  • Knowledge of related rates in calculus
  • Proficiency in algebra and trigonometry
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  • Study the concept of related rates in calculus
  • Explore angular motion equations in dynamics
  • Learn about the kinematics of rigid bodies
  • Review trigonometric identities and their applications in physics
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Students and professionals in physics, particularly those focusing on mechanics and kinematics, as well as educators looking for examples of related rates applications in dynamics.

jasc15
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I have been wrestling with this problem for a few days and it is really frustrating me. The image below shows a bar of length l whose endpoints are constrained to move along the lines shown. Point A moves at a constant velocity to the right and I want to find the angle theta as a function of xA.

I dusted off my dynamics textbook, and there are many examples of this set up in the kinematics chapter, but the only analytical solution shown has the angular acceleration of the bar as a function of theta, alpha and xA.

http://img411.imageshack.us/img411/2960/slidingbarqd2.png
 
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Great, thanks! and here i thought no one was looking at this. I apprecate your work. I guess now it's a problem of related rates to find the angular velocity and angular acceleration of the bar about point A given dx/dt.

Edit: I just noticed an error. in your expression [tex]cos\theta=\dfrac{x_{A}-x_{B}}{l}[/tex] you rearranged it to show that [tex]x_{B}=-\dfrac{l\cdot cos\theta}{x_{A}}[/tex] where it should be [tex]x_{B}=x_{A}-l\cdot cos\theta[/tex]. It's still a bit ugly to play with the algebra and trig, but I'm messing with that now.

Thanks again!
 
Last edited:

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