# B Dumb question about inertia and rotational inertia

#### FactChecker

Gold Member
2018 Award
No. The spinning one will resist rotation more than the non-spinning one. The change of the angular momentums will be equal and opposite, but that will appear as a smaller change of orientation in the spinning one.

#### Graca

Ok, centers of mass will move away at the same speed. The push at the edge will make the spinning one rotate less, the not spinning one rotate more. Thank you.

#### snoopies622

If I have two pipes floating in space and neither one is rotating, and I give one a shove in its center and another an equal shove at an endpoint - perpendicular to the direction of the pipe - they can't have equal resulting translational velocity because in the second case some of the energy invested goes into rotation instead.

#### FactChecker

Gold Member
2018 Award
Pushing on the end with an equal force will cause the end to accelerate more than a push at the center will. Are you sure that you are putting the same amount of energy into the two cases?

#### snoopies622

Well, I'm saying that's a premise of my thought experiment. How do I calculate how much of the energy goes into each component of motion? I took a statics and dynamics course many years ago, but there objects were always fixed at a point - not completely free to move.

(perhaps this is worth a new thread altogether)

Last edited:

#### FactChecker

Gold Member
2018 Award
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

Gold Member
2018 Award
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.

"Dumb question about inertia and rotational inertia"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving