Does moment of inertia equal zero when an object is not spinning?

[flyingpig]

Homework Statement

If an object is not spinning, does that mean the object's moment of inertia is 0? I've always thought that it is. But is it only not spinning when the object is at the moment arm?

 

[Doc Al]

If an object is not spinning, does that mean the object's moment of inertia is 0?

Whether an object is spinning or not has nothing to do with its moment of inertia about some axis, which is a property of its mass distribution. (That's like asking, "If an object is not moving, is its mass 0?")

But is it only not spinning when the object is at the moment arm?

 

[flyingpig]

Let's say I have a pulley spinning on its center (the axis is perpendicular to the pulley) and attached to the pulley is a block of mass.

Clearly, the pulley has a moment of inertia, but does the block too?

 

[Doc Al]

Clearly, the pulley has a moment of inertia, but does the block too?

Sure it does. But who cares, since it's not rotating?

 

[flyingpig]

So it is zero right? That's my question. The object is not rotating, but does that mean the moment of inertia is 0? Even though the radii from the axis clearly is not?

 

[gneill]

So it is zero right? That's my question. The object is not rotating, but does that mean the moment of inertia is 0? Even though the radii from the axis clearly is not?

It seems to me that you're straining to make an analogy between moment of inertia and linear momentum. A better analogy would be between moment of inertia and mass. Mass doesn't "disappear" when velocity is zero, and the moment of inertia doesn't "disappear" when the angular velocity is zero.

Both mass and moment of inertia are properties of things, the former depending upon the total quantity of matter while the latter depends upon that as well as its spatial distribution.

 

[Doc Al]

So it is zero right? That's my question. The object is not rotating, but does that mean the moment of inertia is 0? Even though the radii from the axis clearly is not?

No. The fact that it's not rotating means that its rotational kinetic energy is zero, not that its moment of inertia is zero.

 

[gomunkul51]

You have a misconception about what "moment of inertia" is.
but we can fix that !

The basic analogy is to mass.
Mass is a characteristic of an object that describes its resistance to movement in space, i.e move up/down, sideways and forwards/backwards.

Thus, "moment of inertia" is a characteristic of an object that describes its resistance to movement around a certain axis.

Does an object which is standing still have no mass? obviously it has. because when you star to move it - it resists you according to it's mass (this is by definition of classical mechanics).

Does an object which is not spinning have no "moment of inertia"? obviously it has. because when you star spinning it around some axis - it resists your spin according to it's "moment of inertia" at that axis.

This is the "intuition" behind it.

For more mathematical depth:
http://en.wikipedia.org/wiki/Moment_of_inertia

 

[flyingpig]

What if the block is descending down? Clearly it has a velocity downwards (or same as the "tangenital velocity" in this case).

Treat the block as mr^2

So wouldn't the linear kinetic energy now be mv^2?

 

[gneill]

What if the block is descending down? Clearly it has a velocity downwards (or same as the "tangenital velocity" in this case).

Treat the block as mr^2

So wouldn't the linear kinetic energy now be mv^2?

What do you mean by: "Treat the block as mr²"? In what way and by what laws of physics?

 

[flyingpig]

As in point mass,moment of inertia

 

[Doc Al]

What if the block is descending down? Clearly it has a velocity downwards (or same as the "tangenital velocity" in this case).

Treat the block as mr^2

Seems like a very convoluted way to approach the problem. What's 'r' measured with respect to? It would not be constant as the mass descends. What would be the angular velocity? The radial velocity?

So wouldn't the linear kinetic energy now be mv^2?

No. When all is said and done, the kinetic energy would still be 1/2mv^2, like always.

 

[vela]

What if the block is descending down? Clearly it has a velocity downwards (or same as the "tangenital velocity" in this case).

Treat the block as mr^2

So wouldn't the linear kinetic energy now be mv^2?

What do you mean by: "Treat the block as mr²"? In what way and by what laws of physics?

As in point mass,moment of inertia

Glad you cleared that up. It would really help if you would explain what you're thinking in more detail, even if you think it's obvious what you mean. We can't read your mind, and to us, these little fragments you post, whose meaning may be obvious to you, are often meaningless at worst and confusing at best.I'm still not sure what you are asking, but based on what Doc Al posted, I have a guess. Let me describe a less complicated situation which I think illustrates your basic question:

Let's consider the Earth orbiting the Sun. On the one hand, we can say it moves with speed v at a distance R from the Sun. The kinetic energy of the Earth due to its linear motion would then be given by K_L = ½mv². On the other hand, since the Earth is going around the Sun, we can also say the Earth is undergoing rotational motion. It has an angular speed of ω=v/R, and its moment of inertia would simply be I=mR². The kinetic energy due to this rotational motion is then K_R = ½Iω² = ½(mR²)(v/R)² = ½mv². Therefore, the Earth's total kinetic energy should be K = K_L+K_R = ½mv²+½mv² = mv², but it's not. The kinetic energy is simply K = ½mv². Is that what you're confused about?

 

[flyingpig]

Yes vela! That's a similar way of my question! Sorry for the messy question guys