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. Here is the picture, but my angle is 81.5, http://www.solutioninn.com/physics/mechanics/linear-momentum/in-the-1968-olympic-games-university
One must first transform the value of the angle from degrees into radians
180 degree…………$pi$ rad
81.5 degree ……….x rad
$x =81.5*pi/180 =1.422 radians$
We can compute the radius of the circle from which the arc of length L is a part.
$L = alpha*R$ where $alpha$ is the angle of the arc measured in radians.
$R = L/alpha =L/1.422 =0.703*L$
In the isosceles triangle formed by the two radius that limit the arc of length L the height of the triangle is
$H = R*cos(alpha/2) = 0.703*L*cos(81.5/2) =0.532*L$
The distance of the center f the mass outside the rod is simply (from the symmetry of the problem)
$D = R-H = 0.703*L -0.532*L =0.170*L$
which for $L = 1.80 meter$ is $D = 30.7 cm$ which is consistent with the raise of the world record by about 30 cm.