A stick of length ℓ₀, at rest in reference frame S, makes an a...

Question

A stick of length ℓ₀, at rest in reference frame S, makes an angle θ with the x axis. In reference frame S', which moves to the right with velocity $\vec{v}$ = vî with respect to S, determine (a) the length l of the stick, and (b) the angle θ it makes with the x' axis.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. In reference frame capital fa ladder of proper length L zero is at rest making an angle beta with the Y axis. Now consider frame capital F prime which is moving leftward along the X axis at a velocity of V I hat relative to F I find the contracted length L prime of the ladder as seen in F prime, I, I determine the new angle beta prime that the ladder forms with the Y prime axis as it moves in frame F prime. OK. So it appears for this particular problem we're asked to solve for two separate answers. For part I, we're trying to figure out the contracted length L prime of the ladder as seen in frame F prime. When I say F note that if every single time I say F, we're saying capital F. So in an F prime and for part I, I, we're asked to determine the new angle of beta prime that the latter forms with the Y prime axis that is moving with the frame F prime. OK. So with that in mind, let's read off our multiple choice answers to see what our final an answer pair might be noting that for part IL prime equals sum expression. And I I it states that beta prime is equal to sum expression. So A is L zero multiplied by the square root of sine squared of beta multiplied by one minus V squared divided by C squared plus cosine squared of beta and arc tangent of tangent tangent of beta multiplied by the square root of one plus V squared divided by C squared B is L zero multiplied by the square root of sine squared of beta multiplied by one minus V squared divided by C squared plus cosine squared of beta and arc tangent of tangent of beta multiplied by the square root of one minus V squared divided by C squared. And C is L zero multiplied by the square root of sine squared of beta multiplied by one plus V squared divided by C squared plus cosine squared of beta and arc tangent of tangent of beta multiplied by one plus V squared divided by C square. And finally, D is L zero multiplied by the square root of sine squared of beta multiplied by one minus V squared divided by C squared minus cosine squared of beta and arc tangent of tangent of beta multiplied by the square root of one minus V squared divided by C squared. OK. So first off, let us go over our given variables. So let us note that in frame F, the latter has a proper length L zero and makes an angle beta with the Y axis. So let's make a note of this. So in frame F there, the latter has a proper length of L zero and it forms an angle beta with the Y axis. So this is focusing on the Y axis. So why axis, so do Y axis? Awesome. So and I'll just put Yayd or actually I'll just write it out YXC. OK. So why axis? OK. So now moving right along here and that's our first point. That's important to note now for a second point in frame F prime, it is moving leftward in the X AI direction. And it's important to note that it has a velocity V that is relative to F. OK. So once again, frame F prime is moving leftward in the X axis direction with a velocity V relative to frame F. OK. So now we could start solving for part I. So focusing on part I, so we're trying to solve for the contracted length L prime within frame F prime. So as we should recall and note that sense F prime moves along the X axis with respect to F only the X components of the ladder's length will be affected by the Lorenz contraction. So let's break down the length into its components. So in frame F, the ladder's length L zero is split into X and Y components. Therefore, we can go ahead and write that LX is equal to lo multiplied by sine of beta and LY is equal to L zero multiplied by cosine of beta. Fantastic. So in frame F prime, however, things changed slightly. So the Y component which is LY is unaffected by the motion because as we should recall, since the motion is along the X axis, so therefore, we can go ahead and write that LY prime is equal to L zero multiplied by cosine of beta fantastic. And the X component which is LX undergoes the Lorenz contraction, which we should recall that LX prime. So LX prime is equal to LX multiplied by the square root of one minus V squared divided by C squared fantastic. And then we need to simplify and state that this is all equal to L zero multiplied by sine of beta multiplied by the square root of one minus V squared divided by C squared Fantastic. So now moving right along here, our next step is we need to figure out the total contracted length, which is the L prime value, which is the final value we're ultimately trying to solve for. So as we should recall, a note, the length L prime observed in the F prime reference or frame of reference, I should say is the hypotenuse of the contracted components of LX prime and LY prime. So therefore, we go ahead and write that L prime is equal to the square root of LX prime squared plus LY prime squared. So substituting in our known variables for LX prime and LY prime, we will find that L prime is equal to the square root of L zero multiplied by sine of beta squared plus L zero multiplied by cosine of beta squared. And then when we simplify this value, we will find that this is equal to L zero because we could bring L zero out of the square root multiplied by the square root of sine squared of beta multiplied by one minus V squared divided by C squared plus cosine squared of beta. And that is our final answer. Hooray, we did it. So let's put a box around it because that's our final answer for part I, but we still need to solve for part I I Awesome. So now let's switch gears here to start solving for part I I. So now that we're solving for part I I, we need to figure out the new angle which is beta prime within the frame F prime. So as we should recall and note the angle beta prime within the frame F prime can be calculated as the angle formed by the contracted X prime and YPY prime components of the latter. So first off, let us find we need to solve four tangent of beta prime, which is equal to LX prime divided by LY prime. So Ly prime Fantastic. So now we could substitute in our known expressions for LX prime and Y prime. So when we do that, we will find that tangent of beta prime is equal to L zero multiplied by sine of beta. And then we need to multiply this by and don't forget now we're multiplying by the square root of one minus V squared divided by C squared. Awesome. And then we need to divide this value by L zero multiplied by cosine of beta. And when we simplify this expression, we will find that it is all equal to sign of beta multiplied by the square root of. And since we're running out of room, let's move it down below. Awesome. So this is going to be all equal to sin A beta multiplied by one minus. So one minus V squared divided by C squared fantastic. And then we divide everything by cosine of beta. And that will be our simplified answer. So we can even simplify even further. And we can write that tangent of beta prime is equal to tangent of beta multiplied by the square root of one minus V squared divided by C squared fantastic. So now what we need to do is we need to solve for beta prime. So the angle beta prime in the frame F prime is given by the following equation which states that beta prime is equal to arc tangent of drum roll, arc tangent of beta multiply. So it's the arc tangent of so our tangent of tangent of beta fantastic. And then we need to multiply this by one minus V squared divided by cease work. And that's it we solve for our final answer. Hooray, we did it. So now we can go back to look at our multiple choice answer to see which one corresponds with the answers that we found together. And it appears that the final answer has to be multiple choice answer letter B which states that I is L prime is equal to L zero multiplied by the square root of sine squared of beta multiplied by one minus V squared divided by C squared plus cosine squared of beta. And I I states that beta prime is equal to arc tangent of tangent of beta multiplied by the square root of one minus V squared divided by C squared.

A stick of length ℓ₀, at rest in reference frame S, makes an angle θ with the x axis. In reference frame S', which moves to the right with velocity $\vec{v}$ = vî with respect to S, determine (a) the length l of the stick, and (b) the angle θ it makes with the x' axis.

Key Concepts

Length Contraction

Reference Frames

Relativistic Angles

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A stick of length ℓ₀, at rest in reference frame S, makes an angle θ with the x axis. In reference frame S', which moves to the right with velocity v→\overrightarrow{v} = vî with respect to S, determine (a) the length l of the stick, and (b) the angle θ it makes with the x' axis.

Key Concepts

Length Contraction

Reference Frames

Relativistic Angles

Watch next

A stick of length ℓ₀, at rest in reference frame S, makes an angle θ with the x axis. In reference frame S', which moves to the right with velocity $\vec{v}$ = vî with respect to S, determine (a) the length l of the stick, and (b) the angle θ it makes with the x' axis.