Dadface said:
You asked me to identify the object that has this apparent weight and I thought I had already done this in earlier parts of the thread
...
Specifically I was talking about objects spinning or rotating in horizontal circles.
Fair enough.
Dadface said:
By considering a gyroscope where most of the mass is concentrated in a very thin ring on its outer edge I have derived an equation giving the apparent weight and by plugging numbers into this the apparent weight can be predicted. It is the support force ie the apparent weight that I have calculated.
Again, fair enough. However, if you are referring to this equation ...
Dadface said:
The fractional loss of weight is given by:
f^2=(2v^2.g.r.cos phi)/(v^4+g^2.r^2)
v= speed.
g=acceleration due to Earth's gravity at the particular height of the object/ring.
phi = the angle that the weight vector makes with the orbital plane.
r =radius of circle.
then I disagree with your equation. (BTW, when you say v=speed, I'm assuming that you mean the tangential speed of the ring of the gyroscope.)
Since you politely took the time to clarify these issues, then I will humor your request for a fbd. I model a gyroscope as an infinitesimally thin ring of uniform mass density attached to the outer edge of a massless disk of radius, r, attached to the middle of a massless spindle of length, 2h. Neverminding the issue of stability (which I think is anyway irrelevant to the issue of apparent weight loss), I suppose that the spindle is vertical, and supported on a weight scale, of negligible thickness, on the surface of the Earth. I model the Earth as a perfect sphere of uniform mass density, mass M and radius R.
The following involves some basic notions of calculus. I would begin with an fbd that is symmetric about the vertical, with two differential masses, dm, on the opposite ends of a horizontal massless rod of length, 2r, and this horizontal massless rod would be fastened perpendicularly in the middle to the top end of a vertical massless rod, of length, h (so that I have a T-shaped arrangement of massless rods). The bottom end of the vertical massless rod would be supported by a scale. I assume that the horizontal rod with the two differential masses on its ends is rotating at such a rate that the masses are traveling at speed v into or out of the paper. From this diagram, I could calculate the tension in the left half of the horizontal rod to be:
<br />
d\vec{T}=+\hat{x}\left(v^2r-\frac{GMr}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}\right)dm<br />
and the tension in the right half of the horizontal rod to be:
<br />
d\vec{T}=-\hat{x}\left(v^2r-\frac{GMr}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}\right)dm<br />
Of course you recognize the first term in parenthesis as the centripetal acceleration. The second term that is subtracted is the horizontal component of weight. If you want to compare to your results, assuming that I have understood your definitions correctly, you can replace:
<br />
\frac{r}{\sqrt{\left(R+h\right)^2+r^2}}\rightarrow\cos\phi<br />
Anyway, when I add all horizontal forces on all parts of the ring, I determine that the total horizontal force on the ring vanishes, regardless of v.
<br />
\hat{x}\cdot\vec{F}_{tot}=0<br />
Now, all of that was actually irrelevant, because I don't care what is the horizontal force on the ring anyway. I want to know the force that the scale must apply vertically upward in order to prevent the spindle from accelerating vertically downward. I calculate the vertical component of the weight on each of the two differential masses to be the same:
<br />
dF_{gy}=-\frac{GM\left(R+h\right)}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}dm<br />
In order to integrate this, I will call the linear mass density of the ring, \lambda. After doing the (trivial) integral, I find that the vertical weight component on the entire ring is
<br />
F_{gy}=-\frac{2\pi{}GM\lambda{}\left(R+h\right)r}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}<br />
Since the only other force acting on the system is the support force from the scale, I find that the apparent weight is
<br />
F_{aw}=\frac{2\pi{}GM\lambda{}\left(R+h\right)r}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}<br />
Notice that this is independent of v. This means that I will measure the same apparent weight regardless of how fast the gyroscope is spinning. In particular, I will measure the same apparent weight regardless of whether or not the gyroscope is spinning.
In terms of your notation, my equation says:
<br />
f=0<br />
regardless of v. So, I guess we're stuck.
