# How to calculate the moment of inertia of a bar with two masses on it?

• Davidllerenav
In summary: Ok, once I get that I need to use the Steiner theorem to calculate the moment of inertia of the cylinder with respect to the bar axis, right?Yes.
Davidllerenav
Homework Statement
A bar of length ##L##, width ##a## and mass ##M## revolves around its center, with 2 cylinders of radio ##r## and mass ##m## anchored to each of its ends at a distance d from the center of the bar. Considering the parallel axes theorem, determine the moment of inertia of this system.
Relevant Equations
##I=\int r^2dm##

I first tried to calculate the moment of inertia of the bar. The problem is that I don't understand exactly how are the dimensions of the bar. The fact that it has a width ##a## means that its height is ##a##, or that it has an unknown hight and the width is ##a##, like a parallelepiped? After that I must calculate the moment of inertia of the cylinders with respect to the axis, right? so I can apply the Steiner theorem. The moment of inertia of the cylinders with respecto to their own center of mass is ##\frac{1}{2}mr^2, right? Now, how do I calculate the moment of inertia of the bar in ordder to apply the Steiner theorem?

If you assume that the width is into the page, then that is going to get complicated. It's difficult to tell whether that's what's intended.

But, I think you knew that already.

Davidllerenav said:
The fact that it has a width a means that its height is a, or that it has an unknown hight and the width is a
Since you are given the mass, not the density, the height will not affect the MoI. So it is safe to assume it is horizontal width (into the page). It's not much more complicated than a simple thin rod.
Do you know the formula for the MoI of a rectangle about an axis normal to its plane?

PeroK
haruspex said:
Since you are given the mass, not the density, the height will not affect the MoI.
Yes, I tried to do it that way and I ended up with the same result as it was an one-dimensional bar.
haruspex said:
Do you know the formula for the MoI of a rectangle about an axis normal to its plane?
No, I don't. But I looked on internet, and it is ##\frac {1}{12}M(L^2+a^2)##.

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Davidllerenav said:
Yes, I tried to do it that way and I ended up with the same result as it was an one-dimensional bar.

No, I don't. But I looked on internet, and it is ##\frac {1}{2}M(L^2+a^2)##.

I guess you just use that and, as a separate exercise, derive it if you want to.

It's a nice result. There must be a neat way to prove it. I'll bet @haruspex knows, but he may have gone to bed!

PS I think I see how to do it. It should be:

##\frac {1}{12}M(L^2+a^2)##

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PeroK said:
I guess you just use that and, as a separate exercise, derive it if you want to.

It's a nice result. There must be a neat way to prove it. I'll bet @haruspex knows, but he may have gone to bed!
I'll try to prove it.
PeroK said:
PS I think I see how to do it. It should be:
##\frac {1}{12}M(L^2+a^2)##
Yes, you're right, sorry. What should I do next? How do I use the Steiner theorem in this?

Davidllerenav said:
I'll try to prove it.

Yes, you're right, sorry. What should I do next? How do I use the Steiner theorem in this?

Model the bar as a collection of thin bars spread out across its width and integrate?

jbriggs444
PeroK said:
Model the bar as a collection of thin bars spread out across its width and integrate?
But wouldn't that give me the momenr of inertia of the bar?

Davidllerenav said:
But wouldn't that give me the momenr of inertia of the bar?

If you do it correctly it should give you the MoI of a rectangle.

PeroK said:
If you do it correctly it should give you the MoI of a rectangle.
I should end up with ##\frac {1}{12}M (L^2+a^2)##, right?

Davidllerenav said:
I should end up with ##\frac {1}{12}M (L^2+a^2)##, right?
Yes.

PeroK said:
Yes.
Ok, once I get that I need to use the Steiner theorem to calculate the moment of inertia of the cylinder with respect to the bar axis, right?

Davidllerenav said:
Ok, once I get that I need to use the Steiner theorem to calculate the moment of inertia of the cylinder with respect to the bar axis, right?

The Steiner theorem might be useful before then.

PeroK said:
The Steiner theorem might be useful before then.
Where?

You've got to do some of the thinking for yourself!

PeroK said:
You've got to do some of the thinking for yourself!
I don't know where thw theorem is useful before, but I calculaged the moment of unertia of the cylinder with respect to the center of mass of the bar like this: ##I=I_C+md^2=\frac{1}{2}mr^2+md^2##, thus the moment of inertia of the whole body is ##I_T=\frac {1}{12}M (L^2+a^2)+2 (\frac{1}{2}mr^2+md^2)=\frac {1}{12}M (L^2+a^2)+mr^2+2md^2##. Am I correct?

PeroK
PeroK said:
I guess you just use that and, as a separate exercise, derive it if you want to.

It's a nice result. There must be a neat way to prove it. I'll bet @haruspex knows, but he may have gone to bed!

PS I think I see how to do it. It should be:

##\frac {1}{12}M(L^2+a^2)##
With origin at centre of rectangle, an element dm at (x,y) has MoI dmr2=dmx2+dmy2. So the x and y parts can be integrated separately.

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haruspex said:
With origin at centre of rectangle, an element dm at (x,y) has MoI mr2=mx2+my2. So the x and y parts can be integrated separately.
Can you explain a bit more? How do you get ##mx^2## and ##my^2##?

Davidllerenav said:
Can you explain a bit more? How do you get ##mx^2## and ##my^2##?
Pythagoras; ##r^2=x^2+y^2##.

haruspex said:
Pythagoras; ##r^2=x^2+y^2##.
So ##x## and ##y## are the sides of the rectangle adn ##r## is the diagonal?

Davidllerenav said:
So ##x## and ##y## are the sides of the rectangle adn ##r## is the diagonal?
Yes. If the coordinates of the element are (x,y) then the element is distance r from the centre, so distance r from the axis normal to the plane through the centre. So its MoI about the axis is dm r2.

haruspex said:
Yes. If the coordinates of the element are (x,y) then the element is distance r from the centre, so distance r from the axis normal to the plane through the centre. So its MoI about the axis is dm r2.
I'm not sure that I understand. Is there some picture?

Davidllerenav said:
I'm not sure that I understand. Is there some picture?
Draw a rectangle centred at the origin, this being the axis of rotation.
Mark an element (small patch) at coordinates (x,y). Let its distance from the origin be r.

If the element has mass m, what is its MoI about the axis?
What is r2 in terms of x and y?

haruspex said:
Draw a rectangle centred at the origin, this being the axis of rotation.
Mark an element (small patch) at coordinates (x,y). Let its distance from the origin be r.

If the element has mass m, what is its MoI about the axis?
What is r2 in terms of x and y?
The hypotenuse.

Davidllerenav said:
The hypotenuse.
I mean algebraically.

## 1. How do you calculate the moment of inertia of a bar with two masses on it?

The moment of inertia of a bar with two masses on it can be calculated by using the formula I = (m1 x d1^2) + (m2 x d2^2), where m1 and m2 are the masses of the two objects and d1 and d2 are the distances of the objects from the axis of rotation.

## 2. What is the formula for calculating the moment of inertia of a bar with two masses on it?

The formula for calculating the moment of inertia of a bar with two masses on it is I = (m1 x d1^2) + (m2 x d2^2).

## 3. Can the moment of inertia of a bar with two masses on it be negative?

No, the moment of inertia cannot be negative as it is a measure of an object's resistance to changes in its rotational motion.

## 4. How does the distribution of the masses on a bar affect the moment of inertia?

The distribution of the masses on a bar affects the moment of inertia by changing the distances (d1 and d2) from the axis of rotation. The farther the masses are from the axis, the larger the moment of inertia will be.

## 5. Can the moment of inertia of a bar with two masses on it be calculated if the masses are not evenly distributed?

Yes, the moment of inertia can still be calculated if the masses are not evenly distributed. The formula for calculating the moment of inertia is still applicable, but the distances (d1 and d2) will need to be measured or calculated for each individual mass.

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