How hard must you hit a rod for it to swing up 180 degrees?

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SUMMARY

The discussion centers on calculating the necessary speed at which to hit a thin rod, 0.79 m long, to achieve a complete 180-degree swing over a pivot. Key equations include the change in potential energy equating to the change in rotational kinetic energy, represented as P = K. The moment of inertia for the rod is given by I = 1/12(mL^2), but since the rod pivots at one end, the parallel axis theorem must be applied to find the correct moment of inertia. The user has ruled out speeds of 5.56 m/s and 3.78 m/s as solutions.

PREREQUISITES
  • Understanding of rotational dynamics and energy conservation principles.
  • Familiarity with the parallel axis theorem in physics.
  • Knowledge of moment of inertia calculations for different shapes.
  • Basic proficiency in kinematics, particularly tangential and rotational velocities.
NEXT STEPS
  • Study the application of the parallel axis theorem in rotational motion problems.
  • Learn how to derive and apply the equations for potential and kinetic energy in rotational systems.
  • Explore the concept of rotational inertia for various geometries beyond thin rods.
  • Investigate the relationship between tangential velocity and angular velocity in rotational dynamics.
USEFUL FOR

Students in physics, particularly those studying mechanics, engineers working on dynamic systems, and educators looking for practical examples of rotational motion principles.

Lauren Wright
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Homework Statement


A thin rod, 0.79 m long, is pivoted such that it hangs vertically from one end. You want to hit the free end of the rod just hard enough to get the rod to swing all the way up and over the pivot.
How fast do you have to make the end go?

Homework Equations


P(change in potential energy)=K(change in rotational kinetic energy) = 1/2(I)(Wv)^2
P=mgh (where h is change in height)
I=1/12(mL^2) (the rotational inertia of a thin rod)
V(rotational coordinate omega)=s/r (where s is distance traveled around circle and r is radius)
Wv(rotational velocity)=(change in V)/(change it time)
Wv=(Vt (tangential velocity))/r
Ac (centripetal acceleration)=(V^2)/r
*the change in height of the thin rod is the change in the center, not the tip
*there could be more but I think I covered everything

The Attempt at a Solution


https://goo.gl/rsq8QO <---- url link to a picture of my work
I know the answer is not 5.56 m/s or 3.78 m/s[/B]
 
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Lauren Wright said:
I=1/12(mL^2) (the rotational inertia of a thin rod)

The rod is rotating about one end not the center for which your moment of inertia is for. Do you know the parallel axis theorem?
 

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