Final answer:
The center of mass of a thin semicircular wire of mass m and length L with the origin at the center of the semicircle, which arcs from the +x axis across the +y axis and terminates at the -x axis, is at (0, L/2π). This is determined by integrating over the shape using calculus and recognizing the symmetry of the shape.
Step-by-step explanation:
The question involves finding the center of mass of a semicircular wire with a uniform mass distribution. In problems involving symmetry and uniform mass distribution, the center of mass can often be determined by using symmetry alone. However, in this scenario, the semicircular shape requires the use of calculus to integrate over the shape to find the center of mass. For a semicircular shape of radius R (with equivalent length L for the wire), the center of mass will be along the y-axis due to the symmetry of the shape. The calculation for a semicircular wire yields a position at (0, 2R/π), where R is the radius of the semicircle. In terms of the wire's length L, since the circumference of a full circle is 2πR, the length L for a semicircle is πR. Thus we have (0, (2/π)×(πR/2)) which simplifies to (0, R/2). If we express R in terms of L, we find that R = L/π, substituting this back into our previous answer, we get (0, L/2π), which corresponds to option d.