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Calculating Inertia of object 2

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becma27

Mechanical
May 20, 2002
21
I am trying to figure out the inertia of an object that will be turned by a motor. I want to keep the ratio of inertias below 10:1 and am given that the motor inertia has units of oz-in^2. What kind of inertia is this and how do I calculate it for my object? I have never seen this combo of units and am unclear what I need.

Thanks,
becma27
 
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The units are for mass moment of inertia mass*length^2

TTFN
 
Hi becma27

The motor inertia you quote is what is known as the "moment of inertia" and is derived from summation of (m x r^2)for a rotation of a fixed body.

m=mass of the body

r= distance at which the mass acts from the centre of
rotation.

for example imagine a round weight at the end of a string if you were to wind it round in a circle by hand its "moment of inertia" = mass of weight x (length of string)^2.

Now for your body you have to find the position of the axis of rotation and its mass distribution about that axis, most engineering mechanics books give formula's for various "moments of inertia" of rods, cylinders, squares, rectangles etc.

Also "moment of inertia" is given the symbol (I)

hope this helps
 
The formauls are dependent on what the object is that your turning (Ballscrew, Belt & Pulley, Cylinder, Hollow Cylinder). Each have different equations.

For a solid cylder, parallel witht he shaft, the equations are as follow:

mD^2 Wr^2
J = ---- = ----
8 2g

or antoher

(pi)Lpr^4
J = ----------
2g

J = inertia (oz-in-sec^2)
m = mass (oz/m)
D = Diameter (in)
W = Weight (of cylinder) (oz)
r = radius (in)
g = gravity (386 in/sec^2)
pi = 3.141592564
L = length (in.)
p = material density (oz/in^3)

Now if the shaft was perpendicular to the motor shaft, then the formauls change:

WL^2
J = -------
48g(pi)

(use same legend as above)

Now if you switch to a belt and pulley, again, the formulas change:

Wr^2
Jpulley = -----
2g

J = Inertia (lb-in-sec^2)
W = weight of pulley (lbs)
r = pulley radius (in.)
g = gravity (386 in/sec^2)

To calculate the Inertia of the load:

Wr^2
Jload = -------
g

J = Inertia (lb-in-sec^2)
W = weight of load (lbs)
r = pulley radius (in.)
g = gravity (386 in/sec^2)


It looks like your sizing for servo or stepper based on your 10:1 comment. If your mismatch is higer than 10:1, you will need to add a gearbox. The Inertia seen to the motor is reduced by the square of the ratio.

Jload
Jreflected = ---------
GBratio^2


There are many software packages out there that can help what you want to do. Also, most motion control Mfgs will have an eng. reference in their catalogs with the formuals.

Give us the details (include your motion profile - how far in what time) and I will help you size it.

Cameron Anderson - Sales & Applications Engineer
Aerotech, Inc. -
"Dedicated to the Science of Motion"
 
becma27,

All have provided some useful information to you're question. To get a better understanding of the principle of moments of inertia, I strongly suggest visiting Eric Weisstein's World of Science:


Also a Google search will produce many tutorials and additional insight:


However, the description given by desertfox is incorrect. What is described in that reply is not the moment of inertia, but a moment (torque) exerted by a particle. Many systems have the center of mass and the center of rotation aligned. The system being rotated will still have a moment of inertia. A balanced motor will hopefully have this situation.
I hope all find this information useful.
Best regards,

Matthew Ian Loew

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
Hi MLoew

Yes you are right I am incorrect with my description, I was trying to highlight that the moment of inertia of a body was where the mass of that body was concentrated multiplied by the square of the distance to its centre of rotation
 
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