- #1
frankR
- 91
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An observer, moving at a speed of 0.995c relative to a rod as shown measures its length to be 2.00m and sees its length to be oriented at 30.0 degrees with respect to the direction of motion.
a) What is the proper length of the rod? (Length as measured at rest)
b) What is the orientation angle in a reference frame moving with the rod? (Again, in rest frame)
I've found the correct values, however the math seems ambiguous. I was wondering if there was another way to solve it, with cleaner mathematics.
Do = the length of the rod as measure when the rod is moving a .995c
Lo = the adjacent side of the triangle formed by the rod moving at .995
Lo = DoCos(30.0)
D = the actual length of the rod as observed from the rest frame
L = the actual length of the adjacent side of the triangle formed by the rod in the rest frame
@ = the angle formed between L and D.
h = height of the rest frame triangle
L = Lo/sqrt(1-v^2/c^2) = DoCos(30)/sqrt{(1-(.995c)^2/c^2)}
L = 17.34m
To find D you can use the following equations.
Tan(@) = h/L
Sin(@) = h/D
Cos(@)=L/D
D^2 = L^2 + h^2
I ended up getting this to solve for D.
D = sqrt{L^2 + L^2*Sin^2{Cos^-1(L/D)}^2}
My calculator solved it and got D = 17.26m or 17.42m.
The answers are 17.3m and 3.30 degree.
Is there a cleaner way to do it?
Am I missing something?
a) What is the proper length of the rod? (Length as measured at rest)
b) What is the orientation angle in a reference frame moving with the rod? (Again, in rest frame)
I've found the correct values, however the math seems ambiguous. I was wondering if there was another way to solve it, with cleaner mathematics.
Do = the length of the rod as measure when the rod is moving a .995c
Lo = the adjacent side of the triangle formed by the rod moving at .995
Lo = DoCos(30.0)
D = the actual length of the rod as observed from the rest frame
L = the actual length of the adjacent side of the triangle formed by the rod in the rest frame
@ = the angle formed between L and D.
h = height of the rest frame triangle
L = Lo/sqrt(1-v^2/c^2) = DoCos(30)/sqrt{(1-(.995c)^2/c^2)}
L = 17.34m
To find D you can use the following equations.
Tan(@) = h/L
Sin(@) = h/D
Cos(@)=L/D
D^2 = L^2 + h^2
I ended up getting this to solve for D.
D = sqrt{L^2 + L^2*Sin^2{Cos^-1(L/D)}^2}
My calculator solved it and got D = 17.26m or 17.42m.
The answers are 17.3m and 3.30 degree.
Is there a cleaner way to do it?
Am I missing something?