Rotating rod returning to its initial position

AI Thread Summary
A uniform rod is released and struck vertically at one end to return to its initial position and orientation. The moment of inertia and torque equations are relevant, but the problem is primarily about understanding the motion dynamics rather than energy conservation. The rod undergoes an integer number of revolutions around its center of mass as it moves vertically. The impulse required to return the rod to its original orientation is linked to the initial impulse and the gravitational force acting on it. The discussion concludes with the participant expressing satisfaction after resolving their confusion about the impulse needed.
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Homework Statement


A uniform rod (mass 'm', length 'l') held horizontally above the ground. It is released from its supports and struck vertically upwards at one end. With what impulse should the rod be struck if it is to return to exactly its starting position and orientation?

Homework Equations



I know the moment of inertia of a rod about its centre is ml^2/12,
Torque = Moment of Inertia x Angular acceleration
and the rod moves in the vertical plane around the CM in spirals, but apart from that I'm really stuck.

I can't seem to get the concept of how the rod returns to its starting position.

The Attempt at a Solution


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I tried to conserve energy, mgh=0.5 Iω^2 + KE but I'm not sure where that gets me.

Thanks for any help or pointers :)
 
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This is not an energy conservation situation. Note that the torque about any point on the vertical path of the CM is zero. Also note that in the time the CM goes up and comes back down, the rod undergoes an integer number of revolutions about the CM. Start by figuring out expressions for the initial speed of the CM and the initial angular speed about the CM.
 
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Ahh thank you
So I just found the change in velocity as ωL/6 (from change of angular momentum=radius*linear momentum).
Also that the time taken for it to reach the same place is (2πLn/3g)^0.5 where n is the number of revolutions.
Now how do I get the new required impulse to return it to its original orientation? o_O
How do I know that it actually needs a counter impulse?
 
Helphelphelp said:
... the time taken for it to reach the same place is (2πLn/3g)^0.5
This doesn't look right. The time depends on the impulse J.
Helphelphelp said:
Now how do I get the new required impulse to return it to its original orientation?
What new required impulse? The CM is kicked up by the initial impulse J and is pulled back down by the force of gravity.
 
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can't believe I was so silly, I thought there was a second impulse we had to give the rod doh!
Thanks for your patience, got it sorted now
 
Helphelphelp said:
Thanks for your patience, got it sorted now
OK, what's your answer?
 

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