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Homework Help: Gravitation and a Pendulum

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A pendulum having a bob of mass ##m## is hanging in a ship sailing along the equator from east to west. If the ship sails at speed v what is the tension in the string?. Angular speed of earth's rotation is ## \omega ## and radius of the earth is ## R ##

    2. Relevant equations

    ## T = mg - R\omega'^2##

    3. The attempt at a solution

    If the ship were still, the tension in the string would be given by

    ## T_0 = mg - mR\omega^2##

    However since the ship moves in the direction opposite to earths rotation, the angular velocity of earth with respect to the ship would be ## \omega + \frac{v}{R}##

    Therefore the tension would be
    ## T = mg - mR(\omega + \frac{v}{R})^2 \\
    T \approx T_0 - 2m\omega v
    ##

    Doubt: The answer key says that ## T \approx T_0 + 2m\omega v ## which means the relative angular velocity was taken as ## \omega - \frac{v}{R} ## . But isn't that the relative angular velocity when the ship sails from west to east. Where did I go wrong?
     
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  3. Dec 26, 2017 #2

    Orodruin

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    Here:
    In which direction does the Earth spin?
     
  4. Dec 26, 2017 #3
    The earth spins from west to east.
     
  5. Dec 26, 2017 #4

    Orodruin

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    So if you in addition give yourself a velocity from west to east, how does this affect your angular velocity? Will you get a larger or smaller angular velocity?
     
  6. Dec 26, 2017 #5
    Ohhh yes. Intuitively, I feel it would be a larger angular velocity.
    But if you talk about relative motion, then won't the values get subtracted as they are along the same direction.
     
  7. Dec 26, 2017 #6

    Orodruin

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    Relative to what?
     
  8. Dec 26, 2017 #7
    Angular velocity of the earth relative to the ship.
     
  9. Dec 26, 2017 #8

    Orodruin

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    The angular velocity of the earth relative to the ship is ##v/r##. What you want is the angular velocity of the ship relative to an inertial frame.
     
  10. Dec 26, 2017 #9
    I don't quite get it. It the angular velocity of the earth relative to the ship = ## \omega_\text{earth}-\omega_\text{ship} ##. Say for instance, my frame of reference is that of the ship which is non inertial.
    From the free body diagram, I get the relation ## T = mg - mR\omega^2 ##. But in this case, shouldn't omega be the relative angular velocity of the earth with respect to the ship.
     

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  11. Dec 26, 2017 #10

    Orodruin

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    Yes, and by definition of your problem where the ship is moving at speed ##v## relative to the Earth that relative angular velocity is ##-v/r## so ##\omega_{\rm ship} = \omega_{\rm earth} + v/r##.

    No. It should be the angular velocity relative to an inertial frame.
     
  12. Dec 26, 2017 #11
    Ohh ok. Let me just repeat what I understand from your explanation. Can you please tell me if I understood right?

    Since the ship is on earth the velocity ## v ## given to us is actually its velocity relative to earth. Hence,
    ## -\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R} ##
    And this is ## \omega_\text{ship} ## is to be used in the equation ## T = mg - mR\omega^2 ##
     
  13. Dec 26, 2017 #12

    Orodruin

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    No. You mixed ip the signs again. What I gave you was the relative angular velocity given in post #9 ie earth-ship, not ship-earth. You need to be more careful when you consider those.
     
  14. Dec 26, 2017 #13
    I'm sorry but I'm quite confused right now. If I use

    in ## T = mg - mR\omega^2 ## then I get the result which I posted in the question of this thread:
    However, the answer is ## T \approx T_0 + 2m\omega v ## in the textbook. From their answer I inferred that they must have taken ## \omega_\text{ship} = \omega - \dfrac{v}{R}##

    Sorry for the trouble again!
     
  15. Dec 26, 2017 #14

    Orodruin

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    I thought you were still discussing west to east motion of the ship.
     
  16. Dec 26, 2017 #15
    Oops! My bad. I didn't mention which case I was talking about. Sorry!
    When I wrote the following, I meant the case when the ship was sailing from east to west: -
    ## -\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R}##

    Is it correct now? ​
     
  17. Dec 26, 2017 #16

    Orodruin

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    Yes, since the ship moves in the opposite direction of the Earth's angular velocity, its velocity relative to the Earth's surface will tend to give it a lower angular velocity relative to an inertial frame.
     
  18. Dec 26, 2017 #17
    Thanks a lot. I've understood the solution to this problem now!
     
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