1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Steel Sphere

  1. Aug 6, 2005 #1
    A steel sphere sits on top of an aluminum ring. The steel sphere (a (average
    coefficient of linear expansion) = 1.1*10^-5/C) has a diameter of 4 cm at 0
    C. The aluminum ring (a = 2.4*10^-5/C) has an inside diameter of 3.9940 cm at 0 C. Closest to which temperature given will the sphere just fall through the ring?

    I thought that if I took the a*L*deltaT(aluminum ring) - a*L*deltaT(steel) =
    1*10^-4 (negative difference of diameters, sphere - ring), I would get the
    right answer. However, I wasn't all that close (ended up being 208 C, and I
    got an answer much less than that). I was wondering if there was a better way to approach this problem? Thanks a lot
     
  2. jcsd
  3. Aug 6, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You know the diameters of the aluminum and steel at T = 0. Figure out the diameter of each as a function of [itex]\Delta T[/itex]. Then solve for the temperature at which the diameters are equal.
     
  4. Aug 6, 2005 #3
    This has definitely helped me, but I'm stumped on what to do about the deltaL in thermal expansion equation. And if I do figure that out, what do I do to figure out T?

    Would it be deltaL/L*a(steel) + deltaL/L*a(aluminum) = T?

    Thank you a ton
     
  5. Aug 7, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    The diameter at temperature T equals the original diameter (at 0 degrees) plus the change in diameter from the temperature increase ([itex]\Delta T = T[/itex]):
    [tex]D = D_0 + D_0 \alpha T[/tex]

    Use this to solve for the temperature at which both diameters are equal.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?