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Relativity, space contraction and rotation

  • Thread starter imphat
  • Start date
10
0
1. Homework Statement
A bar (lo = 10mm) is moving along the x axis with speed v according to a referential R.

Meanwhile a table, paralel to the bar is moving along the z axis, with speed u. There's a circular hole in the table with radius = 5mm.

From R perspective the bar has l < 10mm and it fits the hole on the table. But, from the point of view of R', resting in relation to the bar, the hole is smaller than the length of the bar.

A. Does it make sense to ask how the bar fits in the hole? Explain

B. show that in R' the bar and the table are no longer paralel, and theres a angle A between them and tg(A) = uvg/c², g= sqrt(1/(1-v²/c²))

C. show that the point C, in the middle of the whole approaches the point B, in the middle of the bar following the line z'=-u/gv x'



2. Homework Equations



3. The Attempt at a Solution
Im really clueless here. Anyone care to explain this problem to me?
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
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Hi imphat! :smile:

Hint: leave A until you've done B and C.

Can you do B? What have you tried? :smile:
 
10
0
hi tim, thx for the fast reponse once again

the only thing i did so far was to draw some drafts of what happens...

i was thinking maybe rotating the axis on the R' referential, in a way that the movement of the hole is done along 1 single direction, lets say, the z' axis... but i really dont see how the table gets not to be parelel to the bar on R'
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
… avoid doom …

Hi imphat! :smile:

No no no …

In relativity, trying to think up a logical answer usually spells doom.

You have to apply the Lorentz equations!

Just to start you off …

Two corners of the table are at (t,a,0,ut) and (t,b,0,ut) in R (with a and b constant).

So where are they in R´? :smile:
 
Last edited:

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