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Pendulum clock

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A pendulum clock in the centre of a large room is observed to keep correct time. How many seconds per year will the clock gain if the floor is covered by a 1m thick layer of lead of density 11350kgm^-3 ?

    Newtons gravitational constant is G = 6.67 x10^-11 Nm^2 kg^-2

    2. Relevant equations


    3. The attempt at a solution
    Im not sure if its relevant to use but i thought i could use the equation for a pendulum
    T = pi(l/g)^1/2

    Any ideas?
     
  2. jcsd
  3. Nov 19, 2007 #2

    Kurdt

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    It seems this question wants you to work out the new acceleration due to gravity when the lead is added.

    Is that the question in full as you've typed it?
     
  4. Nov 19, 2007 #3
    It is the second part to a question but i didnt think the first part related. The firt part is as follows:

    Show that the gravitational field due to a horizontal uniform thin disc (thickness d, radius R and density p ) at a distance h vertically above the centre of the disc has magnitude

    2piGpd (1 - (h/(R^2 + h^2)^1/2 )

    I was able to do this. But in the second part i dont see that i can use this as it is for a disc and i dont know what shape the room is or how big it is. Can anyone spot something ive missed here?

    p.s thanks for the speedy response
     
  5. Nov 19, 2007 #4

    Kurdt

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    Well since the first part is talking about discs, I'd say you would have to make some assumptions about the room in the second part. Namely that the room is circular and that its radius is a lot larger than the pendulum's height from the disc. If you state the assumptions in the solution then that should satisfy whoever is marking it. You can then work out a correction to g, and apply it in the pendulum period formula from the first post.
     
  6. Nov 19, 2007 #5
    excellent idea, thanks
     
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