Question:
Parallel Axis Theorem?
anonymous
2007-08-09 13:58:39 UTC
I'm in Statics right now and we just had a lecture about area moments of inertia and the parallel axis theorem. I'm still confused about when to use the parallel axis theorem when given a problem to determine the polar moments of inertia. Does anyone have any tips?
Six answers:
gbrb7777
2007-08-09 14:51:55 UTC
This seemed to be a hard concept to grasp in my statics class, although i had no problem with it, or all of statics... Anyway, The professor constantly stressed this point during lectures, but its very simple, I'll try to explain as best as i can..



First, you have to realize that a moment of inertia is a fourth dimensional concept, therefore we cannot directly comprehend what is going on.



Next, you need to know what a centroid/center of mass is and how to calculate it. This is very important, because it leads to the start of the parallel axis theorem.



OK, first an example without using the P-A Theorem. Picture a rectangle 20 feet long and 5 ft wide. now we are asked to find the inertia about the x and y axis. The x axis is located at 10 ft high and 2.5 feet wide.

recall that the inertia of a rectangle is (1/12)bh^3.



Now we must find or simply reason that the moment of inertia of the entire object has the exact same axis as the centroid. Find this using the above formula, and you will see that they are both on the same line of reference. Where the x-axis is placed can change the entire problem. Now I'll do the same problem changing only one thing, the placement of the x-axis.



Same problem as above, but this time the x-axis is at a height

of of 0 feet.



We already know the centroid is at 10ft above the x-axis.But this time it is NOT the same as the reference bar, in this case the x-axis. So this is where the P-A Theorem comes into play.



First picture a figure skater in a tight spin, the reference axis in this case is right down through the very center of the persons head all the way down. Now think about how when they are low and further out from this line they spin slower. But when they get really tall and thin, they start spinning much faster.



The basic concept is that the further out the centroid is from the reference axis that it is spinning about, the higher the inertia is going to be. That is why you add this part of the equation in.



The special edition to the equation is Ad^2, where A is the area and d^2 is the distance from the reference axis to the centroidal axis of the figure at hand.



So, for our problem our 2 axises are at a distance of 10ft apart. The area of the figure we are rotating around the reference axis, is 100 square feet. So Ad^2 is equal to 200. So when calculating the inertia, you must add this part in when the 2 axises are separated. Our sample problem would be the same as the first one without using the P-A Theorem PLUS the new term Ad^2.



Some more hints:

1) Don't try to memorize all the formulas for different shapes. Always break the shape up into easier, less complex shapes, and simply add all of the pieces together. For instance an I beam can be broken into 3 separate pieces. Calculate each one individually and add them all up!



2) It's a really good idea to use a table to keep track of all your calculations, I never had any problems using the tables, its very straightforward.



3)The reference is not limited to where they tell you. It turns out that you will get the exact same answer no matter where it is located. Therefore always try to make it so that a majority of your pieces do not require the use of the P-A Theorem.



That's about all I can think of right now. Hope this helps! Good luck
oberlander
2016-09-29 05:59:27 UTC
Parallel Axis Theorem
David Z
2007-08-09 14:15:31 UTC
One easy way to think about it is in the case where something is spinning, but not at its center of mass.



For instance, if you have a rod, the moment of inertia is 1/12*M*L^2, and this is for a rod spinning at its center. But what if it's spinning at its end? Then the axis (which is parallel to the first axis, meaning they both point in the same direction) is displaced by a distance L/2 from the center. So by the parallel axis theorem, the new moment is 1/12*M*L^2+M*(L/2)^2



Or 1/12*M*L^2+1/4*M*L^2



This is equivalent to 1/3M*L^2, and this is the moment of inertia at for a rod spinning from one end.



Or we can find it for a rod spinning not at its end, but at the midpoint between the center and the end, or at L/4.



1/12*M*L^2+M*(L/4)^2



1/12*M*L^2+1/16*M*L^2



I=1/4(1/3+1/4)*M*L^2



I=1/4(7/12)*M*L^2



I=7/48*M*L^2



It makes these calculations very easy.



Or maybe you have a disk which is being spun around a string a distance 4R from the center of the disk, which has a radius of R. I for a disk is 1/2*M*R^2



But it's 4R away from the real axis.



So



1/2*M*R^2+M*(4R)^2



I=1/2*M*R^2+16*M*R^2



I=33/2*M*R^2



This should give you a good idea about when to use it.
?
2016-03-13 02:31:46 UTC
The total mass of the rod is M so the individual rod centroidal inertia is: (1/12)(M/4) a^2. Also the 'd' term in your transfer should be a/2 since the transfer is 1/2 the square side. All together this gives: I = 4*(1/12)(M/4) a^2 + (M/4) * (a/2)^2 = (1/12 + 1/16) Ma^2 = (7/48) Ma^2
quirky_ki
2007-08-09 14:13:48 UTC
I haven't done this stuff for years so I know my answer is cr*p.

From what I remember you need to find the centroid of the body before you can calculate moments of inertia, so you would use the parallel axis theorem to find the centroid.
usman
2016-03-15 00:14:40 UTC
why the coordinates of the centre of mass of the body always remain constant? If anyone have any concept about this question Plz help me


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