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1 December 2024 20:48
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Question |
| Asked by: |
Harvey Fiala |
| Subject: |
Gyroscopic Precession |
| Question: |
TAKE HOME QUIZ # 3A: Gyroscopic Precession
It is now two months and no one has answered Take Home Quiz # 2 (October 13) or # 3 (October 27). I am a bit disappointed. However, in this Quiz 3A I will give the answer for part three and hope that someone will pursue the answers for parts 1 and 2. For reference, I will repeat the original Quiz # 3 starting in the next paragraph.
Assume the earth were a hollow spherical shell with the thickness of the spherical shell equal to only one foot and the outer diameter of the shell were exactly equal to that of the earth. Assume the density of the shell was uniform and great enough so that the gravitational force at the surface of the shell was exactly equal to the gravitational force (g) at the surface of the real earth. Then from all external appearances, the real earth and the hollow shell would look alive and have identical external gravitational fields. Let us call this spherical shell the “earth shell”.
PART 1. How does the gravitational field strength vary from the outside surface of the one-foot thick shell to its inner surface? Assume the gravitational field strength at the surface of the earth is simply g (32.2 ft/sec2).
PART 2. Calculate the gravitational field strength inside the hollow shell as a function of the radial distance from the center.
PART 3. Assume the earth is the hollow spherical shell described above. Assume that a gyro were undergoing natural precession due to the gravitational field on a pivot point just above the surface of the earth. Let the precessional angular velocity be ωp . Now assume the same gyro were on a pivot point only one foot lower, just at the inside surface of the spherical shell. With everything else being equal except the gyro is now at the inner surface of the shell instead of on the outer surface of the shell, what is the new precessional angular velocity due to natural precession?
ANSWER TO PART 3: The precessional angular velocity on the outer surface of the earth shell was given to be ωp. JUST ONE FOOT DOWN, NEAR THE INNER SURFACE OF THE EARTH SHELL THE PRECESSIONAL ANGULAR VELOCITY WOULD BE EXACTLY ZERO. THAT IS ωp = 0.
Now can someone please answer why the gyroscope will not precess inside the earth shell? The answer is given by the answer to PART 2.
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| Date: |
23 December 2004
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Answers (Ordered by Date)
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| Answer: |
josh - 05/01/2005 22:59:19
| | | read the answer I gave jason. (two questions below yours) I think this is right.
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| Answer: |
Harvey Fiala - 07/01/2005 05:14:56
| | | Josh: Sorry the answer was wrong.
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