Dumb question about inertia and rotational inertia

[FactChecker]

I think that the premise of the OP was an equal force, not equal energy. If you apply the same force over the same time on the end as you would at the center of mass, the end will accelerate more and move farther. That means more work is done, so more kinetic energy is put into the motion.

You can calculate the kinetic energy. In all cases, the center of mass reacts according to $F = mA$. Depending on how long the force is applied, that gives you the amount of linear kinetic energy. The rotational kinetic energy can be calculated using the duration of the force, the torque and the inertial moments.

 

[snoopies622]

I think that the premise of the OP was an equal force, not equal energy.

That's why I was thinking of starting a new thread.

In all cases, the center of mass reacts according to $F = mA$.

Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass, although I don't know how to prove it one way or the other.

Edit: Actually now that I think about it, I'll try approaching it in a collision/conservation of momentum framework instead, perhaps that'll make things clear to me.

 

[FactChecker]

Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass

Intuitively, if you push on one end, that end will move more than if you were pushing at the center. But the other end will move less than if you were pushing at the center. The motion at the center will be the average of the motions at the two ends, which is exactly the same as if you had pushed at the center.

 

[snoopies622]

Yes that sounds credible, thanks FactChecker. Will now make up and work on some specific problems to see what happens.

 

[A.T.]

Intuitively, it seems wrong to me that applying a (force x distance) to the edge of the pipe would result in the same linear kinetic energy as applying the same (force x distance) to the center of mass,...

Applying the same (force x time = impulse) will result in the same linear kinetic energy.
Applying the same (force x distance = work) will result in the same total kinetic energy.