Problems on Length contraction

[chingcx]

Problems on "Length contraction"

Homework Statement

This is a problem in my text. The idea is that a bar is moving with a high speed (say 0.5c) relative to us. We now want to know what will be the length appeared to us if the bar is parallel, perpendicular and 60 degrees tilted. And we are asked when the bar is tilted, how long is it in the co-moving frame.

Homework Equations

L=(Lo)/y where Lo is the proper length

The Attempt at a Solution

First two are very straight-forward, but I can't understand the difference between the last two questions. I tried to resolve the velocity component but still I could not work it out.
Any explanation is appreciated, thank you in advance.

 

[Winzer]

Ok. So you know in this problem that the observer of the bar is going to see the bar a shorter length in the, let's say +x direction. The only parts of the bar that will shrink are the parts parallel to its' velocity; its not going to shrink vertically or diagonally. You will have to modify your Lorentz factor to account for the velocity in the x direction only.

 

[physixguru]

The length of any object in a moving frame will appear foreshortened in the direction of motion, or contracted. The amount of contraction can be calculated from the Lorentz transformation. The length is maximum in the frame in which the object is at rest.

 

[chingcx]

Ok. So you know in this problem that the observer of the bar is going to see the bar a shorter length in the, let's say +x direction. The only parts of the bar that will shrink are the parts parallel to its' velocity; its not going to shrink vertically or diagonally. You will have to modify your Lorentz factor to account for the velocity in the x direction only.

Ok and thank you, but did you misunderstand my question? The bar is still moving in +x direction, but the orientation of the bar is not along +x direction, but make an angle of 60 degrees instead. So I think the velocity of the bar is still 0.5c in +x direction, and this is the thing I don't really understand...

The length of any object in a moving frame will appear foreshortened in the direction of motion, or contracted. The amount of contraction can be calculated from the Lorentz transformation. The length is maximum in the frame in which the object is at rest.

Oh, I know this already, and this problem is somehow different in my opinion, thank you anyway.

 

[Cynapse]

At a just got up guess, you have to resolve the velecity for the x direction (as it is now moving in two planes relative to the stationary reference frame. Also you need to resolve the apparent length in the x direction using simple trig?

 

[Doc Al]

Ok and thank you, but did you misunderstand my question? The bar is still moving in +x direction, but the orientation of the bar is not along +x direction, but make an angle of 60 degrees instead. So I think the velocity of the bar is still 0.5c in +x direction, and this is the thing I don't really understand...

Hint: Only the component of length parallel to the direction of motion will be contracted. Find the x & y dimensions in the proper ("moving") frame, then transform to the "stationary" frame.

 

[chingcx]

Hint: Only the component of length parallel to the direction of motion will be contracted. Find the x & y dimensions in the proper ("moving") frame, then transform to the "stationary" frame.

Thank you. In fact, the textbook problem set v=0.8c (relative to a frame S) and L=1m, and the answers are 0.917m (as observed in the stick's rest frame) and 0.832m (as observed in S) respectively.

From your hint, I get 0.917m, which I expected to be the length observed in S and I've no idea how 0.832m is arrived at. So what's the difference we will see in stick's rest frame and in S frame?

 

[Doc Al]

The length of the stick in its rest frame is 1 m, of course. (That's what L = 1m means, I presume.) The length of the stick in frame S is about 0.917 m. I have no idea what 0.832 m is supposed to be.

 

[richard7893]

Is the bar still at a 60 degree angle as seen in the unmoving frame?

 

[Doc Al]

Is the bar still at a 60 degree angle as seen in the unmoving frame?

No. Since only the horizontal dimension 'contracts', the angle of the bar must change.