Does mass distribution affect an object's ease of rotation?

[Sundown444]

Say we have the moment of inertia of any object. If we removed the radius squared part of the moment of inertia and just leave mass, would heavier, further-from-the-axis mass on one end of the object be as easy to rotate as the lighter mass, maybe closer-to-the-axis of rotation on the other end?

Another question: If moment of inertia is mass times radius squared, counting out the other shapes' formulas, does the radius squared part apply just to one end of the object in question or to all sides of the object?

I hope you all understand what I mean by the first question.

 

[Dale]

If we removed the radius squared part of the moment of inertia and just leave mass

You cannot do this in any sensible manner. The units won’t work out. Since the units of moment of inertia would not be correct then also the units for the follow up question about which is easier to rotate won’t work.

 

[Sundown444]

You cannot do this in any sensible manner. The units won’t work out. Since the units of moment of inertia would not be correct then also the units for the follow up question about which is easier to rotate won’t work.

This is more of a what if question. I know how serious science is compared to that, but if there is no definite answer for that, at least go ahead and answer the second question I asked.

 

[Dale]

Sure:

If moment of inertia is mass times radius squared, counting out the other shapes' formulas, does the radius squared part apply just to one end of the object in question or to all sides of the object?

The radius squared applies for every point on the object. The mass at that point times the square of the radius at that point is the moment of inertia from that point. All of the moments from all of the points of the object are added together to get the total moment of inertia

 

[Sundown444]

Just to make sure, the mass x radius squared for each side or point is a fraction of the total moment of inertia, right?

 

[Dale]

Yes. If you are familiar with calculus then what I am describing is the integral over the whole object: $\int r^2\: dm$

 

[Sundown444]

Yes. If you are familiar with calculus then what I am describing is the integral over the whole object: $\int r^2\: dm$

Erm, what does the f like letter, the d and the m stand for respectively? The m must stand for mass, but what of the others? ∫ stands for integral, right?

 

[Dale]

Erm, what does the f like letter, the d and the m stand for respectively? The m must stand for mass, but what of the others? ∫ stands for integral, right?

It is integral calculus, if you haven’t covered it yet don’t worry about it. What it basically means is break the mass up into an infinite number of little tiny pieces each of mass “dm” multiply each itty bitty mass by the square of its radius (which may be different for each little bit) and add them all together.

 

[A.T.]

∫ stands for integral, right?

Yes, which just means adding up all the infinitesimally small point masses dm (that the object is made of) multiplied by r² (of that particular point mass).

 

[Sundown444]

It is integral calculus, if you haven’t covered it yet don’t worry about it. What it basically means is break the mass up into an infinite number of little tiny pieces each of mass “dm” multiply each itty bitty mass by the square of its radius (which may be different for each little bit) and add them all together.

So, what exactly do you mean by "little tiny pieces of mass"? You mean as in multiple small points each radius covers?

I mean, isn't there a way to cover all the mass at the end of the radius of an object? Say there is a large barbell, and that barbell has one end of it being heavier than the other. Wouldn't mass times radius squared cover that whole end of the barbell?

 

[jbriggs444]

So, what exactly do you mean by "little tiny pieces of mass"? You mean as in multiple small points each radius covers?

Conceptually, you carve the big mass up into little pieces. It turns out not to matter how the individual pieces are shaped as long as they do not overlap with each other and as long as all of them together make up the big mass with no bits left out. [Except possibly for the zero-width boundaries between the pieces]

A "Riemann integral" is what you get when you consider the sum for a set of pieces of a given size and then take the limit as the size goes to zero.

 

[Sundown444]

Conceptually, you carve the big mass up into little pieces. It turns out not to matter how the individual pieces are shaped as long as they do not overlap with each other and as long as all of them together make up the big mass with no bits left out. [Except possibly for the zero-width boundaries between the pieces]

A "Riemann integral" is what you get when you consider the sum for a set of pieces of a given size and then take the limit as the size goes to zero.

But what about this other question of mine?

I mean, isn't there a way to cover all the mass at the end of the radius of an object? Say there is a large barbell, and that barbell has one end of it being heavier than the other. Wouldn't mass times radius squared cover that whole end of the barbell?

 

[Dale]

By the way, when you choose the level of the responses you want “I” indicates you expect college-level responses, which would presume familiarity with calculus. I think you probably really don’t want that level of response. We really can’t teach calculus in a couple of posts.

So here is my last attempt to convey the idea in a non-calculus way: take your large barbell, split it up into tiny 1 mm x 1 mm x 1 mm chunks, wheigh each chunk, and measure the distance of that chunk to the axis, then square that distance, multiply by the mass, and add them all together. That is the moment of inertia.

 

[Sundown444]

By the way, when you choose the level of the responses you want “I” indicates you expect college-level responses, which would presume familiarity with calculus. I think you probably really don’t want that level of response. We really can’t teach calculus in a couple of posts.

So here is my last attempt to convey the idea in a non-calculus way: take your large barbell, split it up into tiny 1 mm x 1 mm x 1 mm chunks, wheigh each chunk, and measure the distance of that chunk to the axis, then square that distance, multiply by the mass, and add them all together. That is the moment of inertia.

Okay, I get it. Thanks!