Rotating rod returning to its initial position

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Homework Help Overview

The problem involves a uniform rod that is released from a horizontal position and struck vertically at one end. The objective is to determine the impulse required for the rod to return to its original position and orientation after being struck.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of energy and the moment of inertia, while questioning the applicability of energy conservation in this scenario. There are attempts to derive expressions for the initial speed of the center of mass and angular speed about the center of mass.

Discussion Status

Some participants have provided insights about the torque and motion of the center of mass, while others express confusion regarding the need for additional impulses. The conversation reflects a mix of interpretations and attempts to clarify the mechanics involved.

Contextual Notes

There are indications of uncertainty regarding the time taken for the rod to return to its original position and the implications of the impulse required. Participants are navigating through assumptions about forces acting on the rod during its motion.

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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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