In the 1968 Olympic Games, University of Oregon jumper Dick Fosburyintroduced a
ID: 1725683 • Letter: I
Question
In the 1968 Olympic Games, University of Oregon jumper Dick Fosburyintroduced a new technique of high jumping called the "Fosburyflop." It contributed to raising the world record by about 30 cmand is presently used by nearly every world-class jumper. In thistechnique, the jumper goes over the bar face up while arching hisback as much as possible, as shown below. This action places hiscenter of mass outside his body, below his back. As his body goesover the bar, his center of mass passes below the bar. Because agiven energy input implies a certain elevation for his center ofmass, the action of arching his back means his body is higher thanif his back were straight. As a model, consider the jumper as athin, uniform rod of length L. When the rod is straight,its center of mass is at its center. Now bend the rod in a circulararc so that it subtends an angle of = 79.5° at the center ofthe arc, as shown in Figure (b) below. In this configuration, howfar outside the rod is the center of mass? Report your answer as amultiple of the rod length L.Explanation / Answer
for simplicity let us take the origin atthe centre of curvature and = 90o
then we get that
L = (1 / 4) 2 r
r = 2 L /
and incremental bit of the rod at angle from the x-axis has the mass given by
dm / r d = M / L
dm = (M r / L) d
yCM = y dm
=(1 / M)=45o=135o rsin (M r / L) d
= (r2 / L)=45o=135o sin d
(onsolving we get)
=(2 / ) [1 - (22 / )] L
=........ L