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maverick280857
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Hi, this isn't exactly homework, as I'm teaching myself, but I thought this forum would be more appropriate for it.
(Page 36, Introduction to Special Relativity by Wolfgang Rindler)
An 18-foot pole, while remaining parallel to the x-axis, moves with velocity [itex](v,-w,0)[/itex] relative to frame S, with [itex]\gamma(v) = 3[/itex], and u, w positive. The centre of the pole passes the centre of a 9-foot hole in a plate that coincides with the plane y = 0. Explain, from the point of view of the usual second frame S' moving with velocity v relative to S, how the pole gets through the (now 3-foot) hole.
[tex]\vec{u} = (v, -w, 0)[/tex]
[tex]\vec{u'} = (u_{1}', u_{2}', u_{3}')[/tex]
Velocity transformation equations:
[tex]u_{1}' = \frac{u_1 - v}{1-\frac{u_1 v}{c^2}} = 0[/tex]
[tex]u_{2}' = \frac{u_2}{\gamma(1-\frac{u_1 v}{c^2})} = -\gamma w[/tex]
[tex]u_{3}' = \frac{u_3}{\gamma(1-\frac{u_1 v}{c^2})} = 0[/tex]
So, [tex]\vec{u'} = (0, -\gamma w, 0)[/tex]
The proper length of the rod is 18ft. Relative to an observer in S', the rod has a length of 9 ft and is moving along the y'-axis with a velocity of [itex]-3w \hat{y}[/itex] ft/s. Also, relative to this observer, the hole is now a 3 ft hole moving along the negative x'-axis with a velocity [itex]-v \hat{x}[/itex].
In the S' frame then, the 9 ft rod is moving down along the y' axis, trying to enter a hole that is moving along the x'-axis in the negative x' direction. This is as far as I've gotten.
How does one explain this observation?
Homework Statement
(Page 36, Introduction to Special Relativity by Wolfgang Rindler)
An 18-foot pole, while remaining parallel to the x-axis, moves with velocity [itex](v,-w,0)[/itex] relative to frame S, with [itex]\gamma(v) = 3[/itex], and u, w positive. The centre of the pole passes the centre of a 9-foot hole in a plate that coincides with the plane y = 0. Explain, from the point of view of the usual second frame S' moving with velocity v relative to S, how the pole gets through the (now 3-foot) hole.
Homework Equations
[tex]\vec{u} = (v, -w, 0)[/tex]
[tex]\vec{u'} = (u_{1}', u_{2}', u_{3}')[/tex]
Velocity transformation equations:
[tex]u_{1}' = \frac{u_1 - v}{1-\frac{u_1 v}{c^2}} = 0[/tex]
[tex]u_{2}' = \frac{u_2}{\gamma(1-\frac{u_1 v}{c^2})} = -\gamma w[/tex]
[tex]u_{3}' = \frac{u_3}{\gamma(1-\frac{u_1 v}{c^2})} = 0[/tex]
So, [tex]\vec{u'} = (0, -\gamma w, 0)[/tex]
The Attempt at a Solution
The proper length of the rod is 18ft. Relative to an observer in S', the rod has a length of 9 ft and is moving along the y'-axis with a velocity of [itex]-3w \hat{y}[/itex] ft/s. Also, relative to this observer, the hole is now a 3 ft hole moving along the negative x'-axis with a velocity [itex]-v \hat{x}[/itex].
In the S' frame then, the 9 ft rod is moving down along the y' axis, trying to enter a hole that is moving along the x'-axis in the negative x' direction. This is as far as I've gotten.
How does one explain this observation?
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