Three identical thin rods, each of length L and mass m, are welded perpendicular to one another as shown 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. (Answer in terms of m and L.)

- for my answer i go t 2Ma^2 but this is not right?

(11/12)mL^2

To determine the moment of inertia of the given structure, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of a body about an axis parallel to and a distance "d" away from an axis through its center of mass is equal to the moment of inertia about the center of mass plus the product of its mass and the square of the distance "d" between the two axes.

In this case, let's consider the rods individually. Each rod can be seen as a thin rod rotating about its center of mass, with its length L and mass m.

The moment of inertia of a thin rod rotating about an axis through its center of mass and perpendicular to its length is given by the formula:

I = (1/12) * m * L^2

Since there are three identical rods, we will calculate the moment of inertia for one rod and then multiply it by 3.

So, the moment of inertia for one rod is:

I_rod = (1/12) * m * L^2

The structure is rotated about an axis that passes through the end of one rod and is parallel to another rod. To determine the distance "d" between these axes, we can consider the geometry of the structure.

Each rod is of length L, and the point of axis rotation is at one end of one rod. If we draw a line from this point to the center of mass of the opposite rod, we can see that this line forms a right triangle with the length of one rod being its hypotenuse. The distance "d" is the length of the other two sides of this right triangle.

By Pythagoras' theorem, we can calculate the distance "d" as:

d = √(L^2 + L^2) = √2L

Now, we can apply the parallel axis theorem:

I_structure = I_rod + m * d^2

Substituting the values we found:

I_structure = (1/12) * m * L^2 + m * (√2L)^2
= (1/12) * m * L^2 + m * 2L

Simplifying:

I_structure = (1/12) * m * L^2 + 2mL
= mL^2/12 + 2mL

Therefore, the moment of inertia of the given structure is mL^2/12 + 2mL.

I cant make out the figure.

for the simplicity let us assume that the rods are thin with the radius much less than L, junction of

the rods as the origin of the coordinates and the axis of rotation as the z axis
for the rod along the y axis we get
I = (1 / 3) m L2
for the rod parallel to the z axis the parallel axis theorem gives
I = (1 / 2) m r2 + m (L / 2)2
= (1 / 4) m L2
in the rod we can see that along the x axis the bit of the material between x and x+ dx has a mass
(m / L) dx
and is at a distance
r = √[x2 + (L / 2)2] from the axis of rotation
so the total rotational inertia will be
Itotal = (1 / 3) m L2 + (1 / 4) m L2 + -L/2∫L/2 [x2 + (L2 / 4)] (m / L) dx
=
I will let you do the calculation.