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I'm a physics major with a quick HW question for a Calc-based physics class. I'll present the problem along with my work. Any help would be appreciated:

Consider 3 uniform thin rods all perpendicular to one another in such a manner that they align with the x, y, and z axii respectively. The centers of mass for the rods coincide with the origin. Each rod has a mass M and a length L. Imagine an axis of rotation parallel to the y-axis running through the very end of the rod along the x-axis. Determine the moment of inertia for the system in terms of M and L.

This appears to be a problem in which the moment of inertia of the system is the sum of the moments of inertia of the individual rods (with respect to the axis of rotation).

For the rod along the y-axis, this is simply calculated from the standard equation of a long thin rod with aor at one end I=1/3 ML^2.

For the rod along the z-axis, this also is an application of the parallel axis theorem. We use the standard equation for the moment of inertia of a long thing rod rotating about an axis passing through its center of mass (perpendicular to the rod's length). This gives I=1/12 M (L/2)^2 +M(L/2)^2. Here, the aor for the entire object is a distance L/2 away from the z-axis (half the distance of a rod).

So far, we have the total inertia of the system to be 1/3 ML^2 + 1/12 M (L/2)^2 +M(L/2)^2

For the rod along the y, no standard form is given, but we know that a single discrete particle has a moment of inertia of MR^2, where R is the radius from axis of rotation. The length of this particle makes little difference if the particle is small. The moment of inertia for a hoop or thin cylindrical shell (different from a hollow cylinder) based on the particle premise is MR^2, where R is the radius from axis of rotation. Summing all our equations:

Algebraically, this reduces to 11/12 ML^2. Is my approach wrong? Apparently, the answer is incorrect.

**HW Problem:**Consider 3 uniform thin rods all perpendicular to one another in such a manner that they align with the x, y, and z axii respectively. The centers of mass for the rods coincide with the origin. Each rod has a mass M and a length L. Imagine an axis of rotation parallel to the y-axis running through the very end of the rod along the x-axis. Determine the moment of inertia for the system in terms of M and L.

*We are looking for an equation that identifies the moment of inertia of the object in terms of M and L.***My Approach:**This appears to be a problem in which the moment of inertia of the system is the sum of the moments of inertia of the individual rods (with respect to the axis of rotation).

For the rod along the y-axis, this is simply calculated from the standard equation of a long thin rod with aor at one end I=1/3 ML^2.

For the rod along the z-axis, this also is an application of the parallel axis theorem. We use the standard equation for the moment of inertia of a long thing rod rotating about an axis passing through its center of mass (perpendicular to the rod's length). This gives I=1/12 M (L/2)^2 +M(L/2)^2. Here, the aor for the entire object is a distance L/2 away from the z-axis (half the distance of a rod).

So far, we have the total inertia of the system to be 1/3 ML^2 + 1/12 M (L/2)^2 +M(L/2)^2

For the rod along the y, no standard form is given, but we know that a single discrete particle has a moment of inertia of MR^2, where R is the radius from axis of rotation. The length of this particle makes little difference if the particle is small. The moment of inertia for a hoop or thin cylindrical shell (different from a hollow cylinder) based on the particle premise is MR^2, where R is the radius from axis of rotation. Summing all our equations:

__Total I__= y + z + x__Total I__= 1/3 ML^2 + 1/12 ML^2 + M(L/2)^2 + M(L/2)^2Algebraically, this reduces to 11/12 ML^2. Is my approach wrong? Apparently, the answer is incorrect.

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