In the 1968 Olympic Games, University of Oregon jumper 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
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
To find the distance outside the rod that the center of mass is located, we need to determine the position of the center of mass for each small segment of the rod and then integrate over the entire arc.
Given that the rod is bent in a circular arc subtending an angle of θ = 81.5°, we want to find the distance outside the rod (let's call it d) in terms of the rod length L.
To integrate over this arc, we can imagine splitting the arc into very small segments with an infinitesimal angle, dθ. The length of each small segment is approximately given by L * dθ.
The center of mass of each small segment will be at a distance r from the center of the circular arc, where r is the radius of the circular arc. In this case, r can be determined using trigonometry:
r = L/2 (since the rod is a thin, uniform rod, the center of mass is at its center)
Now, let's consider a small segment of the rod with a small angle dθ. The center of mass of this segment will be at a distance d from the center of the circular arc. Using trigonometry again, we can relate d to r and dθ:
d = r * sin(dθ)
Substituting the value of r and integrating over the entire arc, we get:
d = integral of (L/2 * sin(θ)) dθ
To find the bounds of the integration, we need to determine the initial and final angles for the arc. In this case, the arc is bent in a circular arc that subtends an angle of θ = 81.5° at the center of the arc. Assuming the initial angle is half of the subtended angle (to make it symmetric), we have:
Initial angle: θ/2 = 81.5°/2 = 40.75°
Final angle: θ/2 + θ = 40.75° + 81.5° = 122.25°
So, to answer your question, yes, when you perform the integration, you should use the bounds of integration from the initial angle of 40.75° to the final angle of 122.25°.
Note: Make sure to convert angles to radians when performing the integration, as most integration functions in mathematical software or calculators expect angles in radians.