Ultra-floppy disk drive operates at speed 60GRPM (f=1GHz).
By how much the proper area of 3.5" floppy disk is increased,
if its diameter remains unchanged ?
2007-05-03 10:20:19 · 6 answers · asked by Alexander 6 in Science & Mathematics ➔ Physics
We can do the integral
A = ∫ 2 π r ζ dr (from 0 to 1.75")
where ζ is the relativistic gamma factor, with the beta at a maximum of 0.933 at the perimeter of the disc. The area comes out to 14.15 square inches, as compared to the non-relativistic 9.62 square inches.
2007-05-03 10:52:25 · answer #1 · answered by Scythian1950 7 · 1⤊ 0⤋
Hmm... wouldn't the area decrease instead of increasing since there will be length contraction?
BTW proper area is area at rest. By definition it is unchanging. The relativistic area would decrease by approximately 73.1% to 7.03 square inches.
2007-05-04 03:20:42 · answer #2 · answered by catarthur 6 · 0⤊ 0⤋
pi times the radius squared
2007-05-03 10:38:18 · answer #3 · answered by Danny 4 · 0⤊ 0⤋
If the diameter does not change, the area does not change.
2007-05-03 10:28:58 · answer #4 · answered by Scott H 3 · 0⤊ 1⤋
Your question makes no sense. I think you missed something out
2007-05-03 10:31:28 · answer #5 · answered by maniacmartinuk 4 · 0⤊ 0⤋
radius times radius times pi
hope i help
2007-05-03 10:30:12 · answer #6 · answered by Anonymous · 0⤊ 0⤋