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# Sizing: Adding a rubber sleeve as a primary roller ## Recap of the last posting

In our last posting, we were comparing the inertia of a 12” long, 0.25” diameter steel rod (0.0013 lb-in^2) to a 1” long, 1” diameter aluminum coupling (0.00957 lb-in^2).

We asked why the lighter weight aluminum coupling would have a larger inertia than a heavier steel rod and the person in the back said “it’s because the coupling’s radius is larger and inertia is a function of the radius squared” and they were correct.

Let’s continue with our calculations by adding another inertial element.

## Adding a roller (rubber sleeve)

I’m going to add a rubber sleeve over the steel shaft. I’m going to make it 10” long. That’s 2” shorter than the steel shaft; because the shaft needs to be supported by bearings on each end and we have to have some shaft length for the coupling to clamp on.

I Googled “density of rubber” and found that hard rubber is 74 pounds per cu-ft. and soft rubber is 69 pounds per cu-ft. This gives us a range of to .

So someone picks a number for us to use.

Do you like ? Great, let’s go with that.

### Calculate the volume of the sleeve

I’m going to pick a circumference that is 2” with an outer diameter of or  0.63662”. Why that value you ask?

Well, one revolution would produce two inches of linear motion if a piece of material was sitting on the surface of the roller. We’ll get into that inertial calculation latter.

Using our trusty volume equation again we get: Multiply the volume by the rubber’s density and we get a weight ( ) of .

Using our inertia equation we get: But wait. Our rubber sleeve has a hole in it that allows it slip over the steel shaft. So let’s calculate that inertia and subtract it from our total.   Subtract from and we get: ### There’s an another way

Now having done all that we could have looked up the inertial equation for a thick-walled hollow tube which is: Where is the inner radius and is the outer radius.

Plug and chug our values: The same as we calculated before.

Note that the weight is its actual weight, by that I mean the weight of the missing center is subtracted from the total (as if it was a solid) weight

### A more efficient way

Now, I don’t know about you, but the first time I looked at that equation I would have said that it should “ ” not “ ”, but you do add the two squared radii together. I’ll leave the derivation of that up to you.

The equation that makes more sense to me is the following: Where is the density and is the height or length of the hollow tube.

This is exactly what we did by using two separate equations. One for the whole rubber sleeve and one for the 0.25” section we removed.

Now that we have calculated the inertia of the rubber sleeve, let’s calculate the total inertia that the motor shaft sees.

Our 1”x 1” coupling has an inertia of .

Our 0.25”x 12” steel rod has an inertia of And our 0.6366” x 10” rubber sleeve has an inertia of Since everything is directly attached to the motor’s shaft, all we need to do is add the three inertias together to get our total inertia of More next time.

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