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Question: Find the proper length of a rod if in the laboratory frame of reference its velocity is \(v = \dfrac...

Find the proper length of a rod if in the laboratory frame of reference its velocity is v=c2v = \dfrac{c}{2} , the length l=1.00ml = 1.00\,m , and the angle between the rod and its direction of motion is θ=45\theta = {45^ \circ }.

Explanation

Solution

We can find the proper length of the rod by using the concept of length contraction of the body moving at an angle with respect to the X-axis which is the length of the body in motion that appears to be contracted. Also, the concept of rest frame and laboratory is used.

Formula used:
The contracted length of the rod by length contraction is given by
l=l01v2c2=l0αl = {l_0}\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} = \dfrac{{{l_0}}}{\alpha }
Where, l0{l_0} - proper length of the rod.
α=11β2\alpha = \dfrac{1}{{\sqrt {1 - {\beta ^2}} }} and β=vc\beta = \dfrac{v}{c}

Complete step by step answer:
Consider the rest frame in which the ends of the rod are expressed in the terms of the proper length l0{l_0}. A(0,0)A(0,0) at origin, B(l0cosθ,l0cosθ)B({l_0}\cos \theta ,{l_0}\cos \theta ) at an angle θ=45\theta = {45^ \circ } at time tt. And in the laboratory frame which is moving with the velocity v=c2v = \dfrac{c}{2} at time tt' are A(vt,0)A'(vt',0) and B(lcosθ1β2+vt,lsinθ)B(l\cos \theta \sqrt {1 - {\beta ^2}} + vt',l\sin \theta ) as the rod is moving along the X-axis, there is no length contraction along Y- axis. So, we can write,
l0cosθ=lcosθ1β2(1){l_0}\cos \theta = l\cos \theta \sqrt {1 - {\beta ^2}} - - - - - (1) and
l0sinθ=lsinθ(2)\Rightarrow {l_0}\sin \theta = l\sin \theta - - - - - - - (2)

Now, squaring and adding equations (1)(1) and (2)(2) , we get
l02=l2[cos2θ(1β2)+sin2θ]l_0^2 = {l^2}\left[ {{{\cos }^2}\theta (1 - {\beta ^2}) + {{\sin }^2}\theta } \right]
Substituting the given values, l=1.00ml = 1.00m and θ=45\theta = {45^ \circ } , we get
l02=12[cos245(1(vc)2)+sin245]l_0^2 = {1^2}\left[ {{{\cos }^2}{{45}^ \circ }\left( {1 - {{\left( {\dfrac{v}{c}} \right)}^2}} \right) + {{\sin }^2}{{45}^ \circ }} \right]
Also, v=c2v = \dfrac{c}{2} , then
l02=[12(114)+12]\Rightarrow l_0^2 = \left[ {\dfrac{1}{2}\left( {1 - \dfrac{1}{4}} \right) + \dfrac{1}{2}} \right]
l02=[38+12]\Rightarrow l_0^2 = \left[ {\dfrac{3}{8} + \dfrac{1}{2}} \right]
l0=1.08m\therefore {l_0} = 1.08m

Hence, the proper length of the rod is l0=1.08m{l_0} = 1.08\,m.

Note: Length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length, which is the length as measured in the object's own rest frame, always note that proper length is greater than the contracted length in moving frame.