A thin (one-dimensional) wire of constant density is bent into the shape of a se

Question

A thin (one-dimensional) wire of constant density is bent into the shape of a semicircle of radius a. Find the location of its center of mass. Assume the origin is at the midpoint of the diameter with the y- axis pointing up. Assume the density is one. Using polar coordinates, determine the integral to be used to most efficiently find the moment with respect to the x -axis, M_x. Use increasing limits of integration. m_x = integral d theta Find the location of the center of mass. On the line of symmetry, the center of mass is located Q units from the diameter. (Type an exact answer.)

Explanation / Answer

Density will be considered 1 .
We will consider a coordinate system. Let the X axis be through vertices of wire, and the y axis be perpendicular to the x axis passing through the halfpoint of the wire

Equation of wire become

y=sqrt(r^2-x^2)) as semicircle

Now calculating

x axis:
x = integral (x dm) / integral (dm)
Converting it into polar coordinates .
x=R costheta, dm = rdtheta
Integrate form theta = 0 to pi.
Mass of the object = * (pi*R)
After integration numerator is zero.
So
x= 0

y axis:
y = integral (y dm) / integral (dm)
Converting into polar coordinates
y=r sintheta, dm = Rdtheta
Integrate form Theta = 0 to pi.
Denominator = Mass of the object = * (r)

After integrating, we find that numerator is 2R^2.
So
y = 2R/pi
So
the center of mass is at (0,2R/pi)

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