Posted by **phys** on Thursday, April 1, 2010 at 3:21pm.

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.

when i do integration, do i use from angle 49.25 to 122.25

- physics(help me with the angle) -
**bobpursley**, Thursday, April 1, 2010 at 5:48pm
You are integrating some distance*dm in order to get a cm (mass*distance). There are hard ways, and easy ways to do do this. By arguments of symettry, you can place the cm on the axis, so the question is where.

consider some dm=k dtheta

but the x part of that, the distance along the final radial, is r-rcosTheta

(draw a diagram to verify that).

So just integrating the arc (-81.5/2 to 81.5/2 deg) will give the position.

Int (r-rcosTheta)dTheta= rTheta+rsintheta over limits, or

limits -.71rad to +.71rad

-r*.71+r(.989)-r(.71)+r(.989)=.558r, or outside the arc, it is .442r.

now in terms of L, L=rTheta=r(1.42) or

r= L/1.42 so

distance is .442*1.42=.62 L

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