Question

In the 1968 Olympic Games, University of Oregon jumper Dick Fosbury
introduced a new technique of high jumping called the "Fosbury flop."
It contributed to raising the world record by about 30 cm and is
presently used by nearly every world-class jumper. In this technique,
the jumper goes over the bar face up while arching his back as much as
possible, as shown below. This action places his center of mass outside
his body, below his back. As his body goes over the bar, his center of
mass passes below the bar. Because a given energy input implies a
certain elevation for his center of mass, the action of arching his
back means his body is higher than if his back were straight. As a
model, consider the jumper as a thin, uniform rod of length L.
When the rod is straight, its center of mass is at its center. Now bend
the rod in a circular arc so that it subtends an angle of θ = 81.5°
at the center of the arc, as shown in Figure (b) below. In this
configuration, how far outside the rod is the center of mass? Report
your answer as a multiple of the rod length L.

y=r^2/L times integral of sin(theta)dtheta
I'm just confused about the angle that i should use, i used angle from 49.25 to 130.75, but i got it wrong.

Answer 1

This is basically an integration problem. You want the CM of a uniform circular arc that faces downward and subtends 81.5 degrees, which is 1.422 radians. Your integral should extand from 49.25 to 130.75 degrees, which is 0.860 to 2.282 radians.

Your integrand appears to be wrong. The difference between the CM and the top of the arc is the integral of
(1 - cos theta)*r*(d theta), divided by the subtended arc angle, 1.42 radians.
the arc length is L = 1.422*r