Maximum height of CM of a rotating stick

AI Thread Summary
A uniform stick is struck at one end while held horizontally, and the problem involves determining the maximum height the center of mass (CM) can rise before returning to its original horizontal position. The discussion highlights the use of conservation of energy, where initial kinetic energy from the strike is equated to potential energy at maximum height. However, it emphasizes that simply considering energy is insufficient, as the stick must complete a half-integer number of revolutions to return horizontally. The relationship between upward motion and rotational motion is crucial for solving the problem. Ultimately, the participants confirm that further exploration of the stick's rotational dynamics is necessary for a complete solution.
Joe8
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Homework Statement


[/B]
A uniform stick is held horizontally and then released. At the same instant, one end is struck with a quick
upwards blow. If the stick ends up horizontal when it returns to its original height, what are the possible values
for the maximum height to which the center rises?

My attempt at a solution is below. I must be doing something wrong. I just do not know what it is.

Homework Equations

The Attempt at a Solution



I tried conservation of energy.

Ei=Ef (i= initial state (right after the strike) f= final state (CM at maximum height))

PEi+KEi(transational)+KEi(rotational)=PEf+KEf(transational)+KEf(rotational)

Taking the origin to be where the stick is initially released => PEi=0
Maximum height the center of mass (CM) can rise to =>KEf(transational)=0
The only external force on the stick is gravity, and gravity exerts no torque on the CM of the stick => Σ[PLAIN]https://upload.wikimedia.org/math/8/1/a/81a69207104f00baaabd6f84cafd15a0.png(external)=0 and Σ[PLAIN]https://upload.wikimedia.org/math/8/1/a/81a69207104f00baaabd6f84cafd15a0.png=dL/dt => dL/dt=0 and L=Iω => ωi=ωf => KEi(rotational)=KEf(rotational)

=> PEi+KEi(transational)+KEi(rotational)=PEf+KEf(transational)+KEf(rotational) will give us
KEi(transational)=PEf
=> 1/2 (m)(vi)^2=mgh

=>h=(vi)^2/2g[/B]
 
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Hello Joe, :welcome:

I suppose then, that just considering energy is not enough !
Joe8 said:
If the stick ends up horizontal when it returns to its original height
you want to translate this into something you can use someway or other: it means it has completed a half integer number of revolutions at this return to original height. The strike gives the stick upward motion plus rotational motion. These two have to be in a certain relationship that you need to explore and exploit.
 
Everything you wrote looks correct, but it is not heading towards an answer.
The key fact is that at the moment it has returned to the original height it is once again horizontal. Think about time and rotation rate.

BvU pipped me, not unusual.
 
Thanks all! I solved it. I will post the answer shortly.
 
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