Maximum height of CM of a rotating stick

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Homework Help Overview

The problem involves a uniform stick that is held horizontally and then released, with one end struck to create upward motion. The question seeks to determine the maximum height to which the center of mass can rise when the stick returns to its original height in a horizontal position.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy and the relationship between translational and rotational motion. There are considerations about the stick's motion and the implications of it returning to a horizontal position after being struck.

Discussion Status

Some participants have provided insights into the relationship between the upward motion and the rotational motion of the stick. There is recognition that simply using energy conservation may not be sufficient to reach a conclusion, and further exploration of the motion dynamics is suggested.

Contextual Notes

Participants note that the stick must complete a half-integer number of revolutions upon returning to its original height, indicating a need to consider the timing and rotational rate in the analysis.

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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