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Simple Pedulum

  1. Nov 9, 2006 #1
    "On the moon of a distant planet, an astrnaut measures the period of a simple pendulum, 0.85m long, and finds it is 4.7 seconds. Back on Earth, she could throw a rock 13m straight up (while wearing her spacesuit). With the same efoort, how far up can she throw the same rock at her present location? Ignore the effects of air resistance."

    I calculated the g on the distant planet to be 1.5191 m/s from using T=2*pi*(square root of L/g).

    Okay, maybe I'm thinking too much for the 2nd part of my work, but I can't seem to figure out how to calculate how high up the ball will travel on the distant planet. It should be higher than 13m because there's less gravitational force acting upon the ball to make it come back down, right?
  2. jcsd
  3. Nov 9, 2006 #2
    The kinematic equations will work the same on the distant planet, I think.

  4. Nov 9, 2006 #3
    Sorry, I just read my note, and it sounds pretty rude. I didn't mean it to come across that way.

    But the idea is to use the kinematic equations. You can use ratios, or calculate the time or whatever.

  5. Nov 9, 2006 #4
    Well I tried to use one of them, and I got 2.016 m as the displacement.

    It just doesn't make sense to me that with using the same amount of force on a different planet with less acceleration due to gravity would make the ball reach a lower height.
  6. Nov 9, 2006 #5
    Can anyone explain please?
  7. Nov 9, 2006 #6
    This question must be split into a few questions

    whats the equation for the period of a pendulum? Does it have the gravitational force in it? Can you solve for the gravitational force with the information supplied in the problem?

    Secondly, how much force did it take to throw the rock that high on the earth? Think potential energy/kinetic energy

    How high would that much force propel the rock the the moon in question?
  8. Nov 9, 2006 #7
    Remember that the energy needed to raise a rock to a level above earth equals the potential energy the rock has at that level.

    edit::I am going to work through the problem to check your solution for "g" because "g" can't be a velocity! It's an acceleration and therefore it's units are l/t^2, l=length t=time.
    Last edited: Nov 9, 2006
  9. Nov 9, 2006 #8
    I'd like to know how you got the correct answer for g without squaring pi and the incorrect units. But anyways, your answer for g is correct. So now just simply use the equation [tex]E_{p}=mgh[/tex] to find out how much energy the astronaught used to throw the rock that high, then using the energy, solve [tex]h=\frac{E_{p}}{mg}[/tex] with the new g to find the height.
  10. Nov 9, 2006 #9
    Question to others, not the OP: What is effort? Does effort mean the same force pushing against the ball, or the same amount of energy expended in throwing the ball? I'll post back after I've had a few minutes to actually think about if these two approaches results in different outcomes.
  11. Nov 9, 2006 #10
    The equation is straight from my book - I only forgot to write out m/s2 instead of m/s.

    Period of pendulum: T=2*pi*(square root of (length of pendulum/g)
    4.7s = 2*pi*(square root of (0.8s/g))
    Square both sides and divide both sides by 4pi^2 to get:
    .5595 = 0.85/g
    g = 1.5191 m/s2

    Anyways, I set the energies equal to each other since I don't know the mass of the rock. g1*13 = g2*x

    Thanks everyone for your help! :)
  12. Nov 9, 2006 #11
    Nope, I was over-thinking the problem... it doesn't matter. The effort can be thought of as the amount of energy transferred to the ball being the same.
  13. Nov 26, 2010 #12
    Try this since you already found the gravitational force of the other planet: multiply the 13m by 9.8m/s^2 to get 127.4. Then divide 127.4 by 1.519 m/s^2 to get 83.87m. I believe this is right maybe someone else could comment on it verifying if its right or not.
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