00:01
So in this problem, we're given the length of the rod, the angle, and the speed of the rod, all relative, or all based on a reference frame moving relative to the rod.
00:12
So here's the example of what the rod looks like.
00:14
You know, you got your two meters, your angle of 30 degrees, and your direction of motion.
00:19
So in order to solve this problem, first we're going to get the x and y likes of this rod in its reference frame currently.
00:25
So that's just going to be l1 times the cosine of 30, which is equal to 1 .73.
00:32
Meters and the y l1y is going to be l1 times sine of 30 and that's equal to one meter and now that you have the x and y values for the original reference frame in order to get to the proper length and proper orientation angle we're going to go ahead and label l2x as what as the x valx component of the proper length so the relation between l1 x and l2 x is just l1 x is equal to l2x divide by the gamma.
01:05
So now let's go ahead and take a little tangent and find gamma.
01:08
So gamma is equal to 1 over the square root of 1 minus v squared divide by c squared.
01:15
So that'd be equal to 1 over 0 .0 .1 the square root of 1 minus 0 .995 squared.
01:24
So 1 over the square root of 1 minus 0 .995 squared, which is equal to 10.
01:32
So here we have found our gamma value...