# Force and Torque Question on an isolated system [duplicate]

If there's a rigid rod in space, and you give some external force perpendicular to the rod at one of the ends for a short time, what happens?

Specifically: What dependence does the moment of inertia have? If it rotates, what is the center of rotation? Does it matter that the rod is rigid? What happens if it's "springy", say a rubber rod instead. Is there a difference between exerting a force for a short period of time, and having an inelastic collision (say a ball hits the end of the rod instead of you pressing).

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The second answer here is helpful: physics.stackexchange.com/q/43232 –  chase lambert Jan 15 '13 at 11:21
What is the center of rotation though? –  chase lambert Jan 15 '13 at 14:15

## marked as duplicate by ja72, Brandon Enright, Dimensio1n0, John Rennie, akhmeteliDec 14 '13 at 7:21

The moment of inertia is the rotational mass of the object and it is solely a function of mass distribution - the shape. Usually it does not matter if it is rotating, but if the rate of rotation is high enough the body could deform. This is a concern in the connecting rods of high speed engines.

The MI does depend on shape, so it will change if the material is not rigid. However, angular momentum will preserved so the object will rotate faster when the MI is smaller.

The center of mass does not change unless acted on by an external force. But that force is being applied from another mass, so an "external" force does not really exist.

Inelastic collisions should be analyzed using the conservation of momentum, and not energy. Then it does not matter the duration of the force, or the strength, just the integrated amount.

Energy is also conserved in the collision, but in an inelastic collision energy is transformed into heat preserving the conservation rule, while providing no help for our solution.

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The moment of inertia of a rod with a perpendicular axis of rotation through the center of mass is $$I=\frac{1}{12}ml^2$$ where m is the mass and l the length.
If you apply a torque $$\tau=r\times F$$ (all vectors), then $$\tau=I\alpha$$ lets you calculate the angular acceleration alpha.