how do you use the parallel axis theorem find the moment of Inertia of Three identical thin rods,?

Question

Three identical thin rods, each of length L and mass m, are welded perpendicular to one another as shown in the figure below. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. Determine the moment of inertia of this structure about this axis. (Use any variable or symbol stated above as necessary.

I don't understand the procedure. You get 11 mL^2/12.

http://www.webassign.net/serpse8/10-p-033.gif


I understand how it works for the y-axis. But I don't know how you use the theorem for the z and x axes.

I have a hard time visualizing the rotation, especially for the z and y axes. It rotates about the y-axis.

Answer 1

For this problem, it is probably best to FIRST find the combined rotational inertia about the y0 axis that goes through the center of mass...AND THEN, use the parallel axis theorem to translate the entire thing to an offset to the y1 axis.



The y-axis rod is infinitesimally thin. It contributes zero to the rotational inertia about the y0-axis.



The x-axis rod is rotated perpendicular to its length, and so is the z-axis rod. These each contribute 1/12*m*L^2



Hence, the answer for the total rotational inertia ABOUT THE CENTRAL y-axis is:

Iy0 = 1/6*m*L^2





OR, if we want to express it in terms of total mass (capital M), realize that:

M = 3*m



OR

m = M/3



Thus:

Iy0 = 1/18*M*L^2





Now, we need to use the parallel axis theorem to translate FROM the center of mass to the edge point of the x-axis rod. We are translating by distance d = L/2.



NOTE OF CAUTION: when using the parallel axis theorem, it is ONLY valid if one of the points involved is the center of mass. The way that it is written in the textbook, it is written for translating to an axis parallel that is FROM the center of mass. You can algebraically reverse the formula to go TO the center of mass. BUT, don't just use it to go between any arbitrary points.







Use of the PAT:

Iy1 = Iy0 + M*d^2



Thus:

Iy1 = Iy0 + M*L^2/4

Iy1 = 1/18*M*L^2 + M*L^2/4



Factor out M*L^2:

Iy1 = M*L^2 * (1/18 + 1/4)



Add the fractions:

1/18 + 1/4 = 11/36



Thus:

Iy1 = 11/36 * M*L^2





OR, as they want you to express it, in terms of individual rod mass:

Iy1 = 11/36*(3*m)*L^2





Conclude:

Iy1 = 11/12 * m*L^2

Answer 2

The answer should be mL^2 / 6 i.e sum of mL^2/12 and mL^2/12 for z and y axis as they are same for x axis because the interchange their position after a 90 degree rotation. For x axis M.I. is 0 as whole of mass lies on the same axis.



You don't have to use parallel axis theorem in this problem, its used only when we have to calculate M.I about some other axis parallel to axis passing through center of mass or vice-verca .

Answer 3

Find the center of gravity of the three rods. find its distance from the axis of rotation. Multiply the total mass of the rods by that perpendicular distance squared and find your moment of inertia.

Answer 4

find the original moment of inertia

then

add it with mass of rod * distance from COM of rod you want to find inertia ^2

for x axis,

you should be given formula for rod at center of rod