Horizontal Force on Pivot Point

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SUMMARY

The discussion focuses on calculating the horizontal force on a pivot point in a physics problem involving torque and angular momentum. The key equations referenced include torque (T=F*r), the relationship between force and angular acceleration (F*X=Ia), and the moment of inertia (I=ML²/3). The user successfully solved part A but struggled with part B, specifically in deriving the equation for the force at the pivot point when the distance X is 0.5L, which corresponds to the center of mass.

PREREQUISITES
  • Understanding of torque and its calculation using T=F*r
  • Familiarity with angular momentum and its relationship to force
  • Knowledge of moment of inertia, specifically I=ML²/3
  • Basic principles of mechanics, particularly regarding pivot points and forces
NEXT STEPS
  • Study the concept of "impulsive torque" and its application in dynamics
  • Learn how to derive equations for forces acting on pivot points in rigid body dynamics
  • Explore examples of calculating angular momentum in various mechanical systems
  • Review the principles of equilibrium in static and dynamic systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to torque and angular momentum.

Epif
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Homework Statement


http://img151.imageshack.us/img151/6641/torquebk4.jpg


Homework Equations



T=F*r
F*X=Ia
I=ML2/3

The Attempt at a Solution


I was able to do part A easily enough, but I'm not quite sure where to even start with part B. I know that the force on the pivot point would be F0 if X was .5L, ie the center of mass. However, I am unable to come up with the equation asked for. Thanks for any help!
 
Last edited by a moderator:
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Hi Epif! :wink:

Hint: take moments about any point on the rod (might as well use the c.o.m.), and use "impulsive" torque = change of angular momentum. :smile:
 

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