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 ?
2007-05-04 07:02:29 · 4 answers · asked by Scythian1950 7 in Science & Mathematics ➔ Physics
tinker, on the nose
2007-05-04 07:48:06 · update #1
Catarthur, the problem presumes that somehow the 3.5" diameter is maintained, but to observers riding on the disc (never mind accelerated relativistic frames), the circumferences appear longer than they appear to us.
2007-05-04 16:19:39 · update #2
How about:
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.
Or are you just wondering why Alexander says "nobody
apparently still gets what is going on here" ?
2007-05-04 07:21:02 · answer #1 · answered by tinkertailorcandlestickmaker 7 · 1⤊ 0⤋
A=3.14159265*r ^2, where r is the radii.
If the diameter remains unchanged, so does the area.
A is not a function of speed.
2007-05-04 14:12:28 · answer #2 · answered by chanljkk 7 · 0⤊ 0⤋
Am I making a dumb mistake by saying that the surface will be reduced? It is lorentz contraction we are talking about, yes?
And proper means at rest, yes?
2007-05-04 22:34:34 · answer #3 · answered by catarthur 6 · 0⤊ 0⤋
The comprehensive examination covers the field of study as determined by the department
and are developed and administered within the department.
2007-05-04 14:04:58 · answer #4 · answered by Joe Red 4 · 0⤊ 2⤋