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2 more thermal expansion questions

  1. Sep 19, 2005 #1
    1.) a steel rod is 3 cm in diameter at 25 C. a brass ring has an interior diameter of 2.992 cm at 25 C. at what common temperture will the ring just slide into the rod?

    for this Q, i am not too sure which equation to use. i think it's the volume change one, which is delta V= beta(Vo)(delta T)

    but i dont know how the temperature and the diameter is going to fit in...

    2.) a grandfather's clock is calibrated at a temperature of 20 C. the pendulum is a think brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 C, does the clock run fast or slow? how much time does it gain or lose in a 24 hour period?

    this Q...i just have no idea how to start.....

    thanks in advance, any suggestions/comments are greatly appreciated.
     
  2. jcsd
  3. Sep 19, 2005 #2

    Doc Al

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    Staff: Mentor

    Forget about volume; the rod will just slide into the ring when their diameters are the same. You need to find the [itex]\Delta T[/itex] that will have them end up with the same diameter.
    Hint: How does the period of a pendulum depend on its length?
     
  4. Sep 19, 2005 #3
    for number one, are u suggesting taht i should use the linear equation instead of the volume one?

    linear equation being delta L=alpha(Lo)(delta T)
     
    Last edited: Sep 19, 2005
  5. Sep 19, 2005 #4
    "Hint: How does the period of a pendulum depend on its length?"

    taht the thing i dont no...so i cant do the Q....
     
  6. Sep 19, 2005 #5

    Doc Al

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    Staff: Mentor

    Absolutely.

    Regarding the pendulum: Look it up! :smile:
     
  7. Sep 19, 2005 #6
    thanks alot.
     
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