Question

The assembly consists of four rods cut from the same bar stock. The curved member is a circular arc of radius b. Determine the y- and z-coordinates of the centroid of the assembly.

Comments

Insufficient information. Are the 4 rods of equal length? How are they assembled? In what arrangement? Do the rods form the circumference of a circle? Does the "curved member" describe one of the rods, or the rod from which the 4 rods have been cut? What is the length of the rod from which the 4 rods have been cut? What is the significance/implication of the y- and z-coordinates---is there an x-coordinate? What is the frame of reference? Are the rods assembled into a 3-dimensional shape (for example, a pyramid with curved edges)? Please provide a picture or more information.

Answer 1

With limited information this solution has made a number of assumptions:

• The rods have been cut from the circumference of a circle radius b, so that each quadrant arc length is πb/2
• The 4 rods are identical
• Two rods form the base of a 3-dimensional figure
• These two rods are joined to one another at their ends
• One end of each of the remaining two rods is joined perpendicularly to the point where the two rods forming the base are joined
• The free ends of the two rods are joined together 

In the x-y plane the equation x²+y²=b² is a graph of a circle with its centre at the origin (0,0) and radius b.

Consider the arc in the first quadrant. Let the ends of the arc be A(b,0) and B(0,b), so the midpoint of AB=X(b/2,b/2). This is also the centroid of the double-arc base, because of symmetry.

Next we need to look at the vertical part of the structure built on the chord AB. We can make X(b/2,b/2) the origin in a different frame of reference. By doing so the new origin will be the centroid of the base and we can find the centroid of the 3D structure which will be the vertical measure (the z-coordinate). The x- and y-coordinates will both be zero in this redefined frame of reference.

If we also tilt the new frame of reference by 45°, then the chord AB will be horizontal in this frame of reference.  Since X is the midpoint of AB, AX=BX and this length will establish the coordinates of x and y in the new frame. AX=BX=√((b-b/2)²+(0-b/2)²)=b√2/2, making A(b√2/2,0) and B(-b√2/2,0).

The two rods forming the vertical part of the structure rest on this new horizontal base AB. One end of one rod will be at A and one end of the other rod will be at B. Their other ends join at Y, and we need to establish the z-coordinate for Y.

One way to do this is to find the intersection of two circles (each with radius b), one with centre displaced from A by radius b and the other with centre displaced from B by radius b. Since AB lies on the horizontal (x) axis, the centres of the two circles are (b√2/2-b,0)=(½(√2-2)b,0) and (½(2-√2)b,0).

The equations of the circles are (x+½(2-√2)b)²+y²=b² and (x-½(2-√2)b)²+y²=b². We already know by symmetry that x=0 (which is a solution of this system of equations), so y²=b²-¼(2-√2)²b².

y²=b²-¼(6-4√2)b²=b²(1-3/2+√2)=b²(√2-½), y=b√(√2-½). (We only need the positive square root.)

In the revised frame of reference this gives us point Y(0,b√(√2-½)).

To find the centroid of this vertical structure we need the length of either of the truncated arcs AY/BY. For this we need to calculate the congruent angles YBA or YAB. If θ is this angle then tanθ=OY/OB=OY/OA where O is (0,0) in the revised frame of reference.

OY=b√(√2-½), OA=OB=b√2/2. tanθ=√(√2-½)/(√2/2)=√(2√2-1)=1.3522, θ=0.9340 radians approx (about 53.52°), making the arc length 0.9340b. If the line density of the rod is ρ then the mass of each arc (AY=BY)=0.9340bρ. Mass is θbρ.

The angle bisectors of angles YBA and YAB meet at Z, the centroid of the two arcs on the vertical (z) axis. Remember that the horizontal axis is in fact the x-y plane in the revised frame of reference.

z=OZ=OAtan(θ/2)=(b√2/2)tan(0.9340/2)=0.356532b approx. We'll just use z=btan(θ/2)/√2. This puts the centroid of the two arcs at (0,0,btan(θ/2)/√2) in 3D. We need the centroid of the whole assembly. The mass of the two vertical arcs is 2θbρ and of the base it's 2(πb/2)ρ=πbρ. If the centroid of the whole assembly is at z=c (x=y=0) we can take moments about c to find z:

2θbρ(btan(θ/2)/√2-c)=πbρc, θbtan(θ/2)√2-2θc=πc, θbtan(θ/2)√2=c(π+2θ), c=θbtan(θ/2)√2/(π+2θ).

So the centroid is at (0,0,θbtan(θ/2)√2/(π+2θ)) making the y-coordinate 0 and the z-coordinate θbtan(θ/2)√2/(π+2θ), in the revised frame of reference.

To apply this formula in a "real" case, substitute for b and use θ=arctan(√(2√2-1)) and substitute for θ.

This is the proposed solution for an assembly of 4 rods into a type of pyramid structure. The problem may of course intend to be associated with a different structure, but there's insufficient given information to determine the intended dimensions of each rod. Although I've provided what I believe to be a solution, there may be errors.