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

AI Thread Summary
To determine how hard to hit a thin rod (0.79 m long) for it to swing up 180 degrees, one must consider the change in potential energy and rotational kinetic energy. The relevant equations include the potential energy formula P=mgh and the rotational inertia I=1/12(mL^2). The rod's moment of inertia must be calculated based on its pivot point, which is at one end, not the center. The discussion highlights the importance of the parallel axis theorem in calculating the correct moment of inertia for this scenario. Understanding these principles is crucial for solving the problem accurately.
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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