Consider a 1m stick on the frame of reference F. The stick is placed on the xy p

Question

Consider a 1m stick on the frame of reference F. The stick is placed on the xy plane, and makes an 37 degrees angle with the x-axis. The frame of reference F moves another frame of reference F at a speed v which is parallel to the x-axis.
a) What is the speed of the meter stick measured 45 degrees to the x-axis on F'?
b) What is the length of the meter stick on F'?
(sorry about the bad English, the textbook I used has been translated poorly and it's really difficult to translate it back to English)

Explanation / Answer

I think what this problem is getting at is length contraction. This is a classic exam of the phenomenon (the contracting of a meter stick to increase the angle it makes with the x axis).

Length contraction only occurs for lengths in the direction of motion and the formula is given by

L = Lp/, where = 1/(1 - (v/c)^2), L is the length when it is moving with speed v and Lp is its rest length (or proper length).

a) So, our meter stick has two length components, one in the y direction (which does NOT change, since the velocity is along the x axis), and one in the x direction, which is the direction along which length contraction occurs. The length along the y axis is just (1m)sin(37) = 0.6m, and the length along the x axis is (1m)cos(37) = 0.8m. If the angle were to change to 45 degrees, then the resulting x and y components would have to be the same length. Since the y value DOES NOT change, the x value will change to equal that of the y value. So,

L = Lp/, or = Lp/L = (0.8m)/(0.6m) = 1.33

This means, 1/(1 - (v/c)^2) = 1.33

(v/c)^2 = 1 - (1/1.33)^2 = 0.44

So, v = (.44)*3*10^8 m/s = 2*10^8 m/s, or 66% the speed of light.

b) If each component is of length 0.6m, then the total length is (2)*0.6m = 0.85m

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