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The ends of three identical uniform thin rods are joined at right angles to form a U-shaped body. A freely rotating axle $A$ is attached to one end of this body. The axle is then fixed to the ceiling such that the body hangs freely from one end. Assume that the system is at equilibrium. enter image description here

Part A) Draw a free-body diagram of the U-shaped body, and deduce the angle made by its upper leg with the vertical.

Part B) Let each thin rod in the U-shaped body have mass $m$ and length $l$. Find the moment of inertia of the body about the axle $A$. Assume that the axis of rotation is perpendicular to the plane defined by the body

Working -

With part A, when the U-shaped rod is in equilibrium, I suspected that the centre of mass of the rod would move such that it would be along the perpendicular of the axle $A$. From there I really am unsure on how to determine the angle it will make. With Part B, I do know that I can use the Perpendicular axis theorem but that's where I get stuck as well.

Any suggestions, directions, advice or even general information about how to tackle these problems would be greatly appreciated. Feel free to edit my question for clarity. Thank you

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closed as off-topic by Gert, user36790, ACuriousMind, John Rennie, Kyle Kanos Nov 17 '15 at 11:16

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First of all, I suppose that "equilibrium" in this case means "static equilibrium" in that the thing is not swinging. You suspect it will be "along the perpendicular of the axle A", but there is not just one perpendicular to the axle. Any line in the plane of the rods is perpendicular, so I also suppose you mean the vertical perpendicular, i.e. straight down parallel to the force of gravity. Further, I assume the "upper leg" will be the currently-vertical rod attached at A as pictured, since in static equilibrium it will be higher than the other rods since they will swing lower.

Continue with the center of mass idea. Each thin rod's own center of mass is of course in the center of the rod. Define your coordinate system, probably using the origin at A and find the distance to the center of each rod, then find the center of mass of the combined rods using the basic center of mass formula. Once found, draw the line from the combined center of mass to the point A. What is the angle that makes with the vertical perpendicular? That's the same angle that will occur if the "upper leg" rotates so the center of mass will be on the vertical perpendicular.

For part B, you don't necessarily need the perpendicular-axis theorem. Yes, the axis is perpendicular to the plane of the rods, but don't let the similar words confuse you. To use the perpendicular-axis theorem, you would need to have the moments of inertia for two other orthogonal axes IN the plane. The just complicates it. (Of course, if you already have the other two moments of inertia formulas, then the perpendicular axis theorem is simple... it would just be the sum.) Instead, find the formula for the moment of inertia of a thin rod about an axis perpendicular to the rod. I found it easily by an internet search. It's probably in most text books covering this subject.

The key in either case will be the parallel axis theorem: $ I = I_{cm} + m d^2 $ where $m$ is the mass of a single rod and d is the distance from the center-of-mass axis to a parallel axis. Note that the center-of-mass axis is referring to an axis through the center of mass of a single rod (not the combined center of mass mentioned above). In other words, d is the distance from point A to the center of a particular thin rod.

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