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

• Sundown444
In summary: So, the limit would be the sum of all the individual dm*r2 for all the points on the object.So, 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] 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
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.

Sundown444 said:
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.

Dale said:
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.

Sure:
Sundown444 said:
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

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

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

Dale said:
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?

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Sundown444 said:
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.

Sundown444 said:
∫ 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).

Dale said:
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?

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Sundown444 said:
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.

jbriggs444 said:
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 said:
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?

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.

Dale said:
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!

## 1. Does the shape of an object affect its ease of rotation?

Yes, the shape of an object can affect its ease of rotation. Objects with a more compact and symmetrical shape tend to have a lower moment of inertia, making them easier to rotate. On the other hand, objects with an irregular shape or distributed mass have a higher moment of inertia, making them more difficult to rotate.

## 2. How does the distribution of mass affect an object's ease of rotation?

The distribution of mass plays a significant role in an object's ease of rotation. Objects with a more concentrated mass distribution have a lower moment of inertia and can rotate more easily. In contrast, objects with a distributed mass have a higher moment of inertia and require more force to rotate.

## 3. Can changing the mass distribution of an object affect its rotational speed?

Yes, changing the mass distribution of an object can affect its rotational speed. By redistributing the mass of an object, the moment of inertia can be altered, which in turn affects the object's rotational speed. A more compact mass distribution will result in a lower moment of inertia and a higher rotational speed, while a distributed mass will result in a higher moment of inertia and a lower rotational speed.

## 4. How does the distance of mass from the axis of rotation impact an object's ease of rotation?

The distance of mass from the axis of rotation, also known as the radius of gyration, affects an object's ease of rotation. Objects with a smaller radius of gyration have a lower moment of inertia and are easier to rotate. Conversely, objects with a larger radius of gyration have a higher moment of inertia and require more force to rotate.

## 5. Is there a relationship between an object's mass and its ease of rotation?

Yes, there is a relationship between an object's mass and its ease of rotation. Generally, the more massive an object is, the more force is required to rotate it. However, the distribution of mass also plays a significant role in an object's ease of rotation, as a more distributed mass can result in a higher moment of inertia and make rotation more difficult.

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