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Angular velocity- pendulums and ideal springs

  1. Jan 24, 2009 #1

    Suppose you were kidnapped and held prisoner by space invaders in a completely isolated room, with nothing but a watch and a pair of shoes (with shoelaces of known length). Explain how you might determine whether this room is on earth or on the moon.

    2. Relevant equations

    ang. velocity (W) = square root of (g/L)

    W= 2 pi (f)

    3. The attempt at a solution

    Ok- I know that gravity is less on the moon than here on earth, so the frequency would be slower. I know that this has something to do with comparing frequency's of the watch and of a pendulum (dangling shoes or watch (?) from shoelaces) but I'm not making the connection on how a person would do this.
  2. jcsd
  3. Jan 24, 2009 #2
    Insert the value for gearth and calculate the what the angular frequency would be (hope you have a pencil in your pocket). Find the period of one cycle from the angular frequency. Construct the boot pendulum with length L, pull the boot about 10 degrees off the vertical and release. Use the watch and time ten cycles. Divide by 10 to give the time for one cycle. If this time agrees with the earth calculation then you didn't leave the earth.
  4. Jan 24, 2009 #3
    Why 10 degrees? I know all of this relates back to the angle that the path swings out somehow. Would it be the same if I did say, a 45 degree angle (easier to meausre) and then counted 45 cycles and divided by 45? What does the angle have to do with it??
  5. Jan 24, 2009 #4
    The equation you are using to determine the angular frequency is for small angles of less than 15 degrees. For larger angles the math is much more complicated to solve the equation of motion and the angular frequency equation becomes dependent on the starting angle which you can't measure unless you walk around all the time with a protractor! Timing for 10 cycles has nothing to do with starting the boot at 10 degrees. I suggested 10 cycles to get a reliable value for the period. You could time for more cylces like 45 then divide by 45 and you will still get the time for one cycle but it would be a little more accurate than timing 10 cycles.
  6. Jan 24, 2009 #5
    Awesome- very helpful! Thanks so much.