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Acceleration due to gravity at centre of hemisphere

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data

    find the acceleration due to gravity at the centre of a solid hemisphere.
    2. Relevant equations

    ##F=\frac{GMm}{r^2}##
    3. The attempt at a solution

    i decided to go for cylindrical coordinayes (which is way beyond my syllabus). I did some research though.
    let me take a point P(r,θ,z) inside the sphere amd an elemental volume dV at P. This P exerts a force dF. But there is also a point Q(r,θ+180°,z) which cancells out the horizontal component of dF. let the line OP make an angle φ with the Z axis. (assuming the hemisphere lies on the xy plane with centre at O)
    ##dg=\frac{Gdm}{r^2+z^2}cos\phi##
    ##dg=\frac{G\rho dV}{{(r^2+z^2)}^{3/2}}##
    now dV=dz.dr.rdθ
    Is this expresion for dV true for all cases? how do you get that expression for dV?
    (i found that expression on some video).

    now i have to integrate the expression.
    so $$g=\int_{\theta=0}^{2\pi}\int_{r=0}^R\int_{z=0}^?f(r,\theta,z)dz.dr.rd\theta$$
    im finding it difficult to find upper limit for z.
     
    Last edited: May 16, 2015
  2. jcsd
  3. May 16, 2015 #2
    Why not use spherical coordinates [itex](r,\theta,\varphi)[/itex] instead? It's a lot easier to get the integration limits in that case.
     
  4. May 16, 2015 #3
    i will try. but what about the expression for dV in cylindrical coordinates and the upper limit for z?
     
  5. May 16, 2015 #4
    Well since the equation of a hemisphere is [itex]0<x^2+y^2+z^2<R^2[/itex] with [itex]z>0[/itex], if you use the fact that [itex]x^2+y^2=r^2[/itex], you get [itex]0<z<\sqrt{R^2-r^2}[/itex] where r is a variable, and R is a constant (the radius). Also, the expression for dV is correct, and is only valid in cylindrical coordinates. For more general coordinates you need to calculate the Jacobian of transformation, here's a good start, see the "Examples" section.
     
  6. May 16, 2015 #5
    why is it dz.dr.rdθ? is it the volume of a small cylinder?
     
  7. May 16, 2015 #6
    Basically it's the volume of an infinitesimal cylindrical shell with mean radius r, thickness dr and height dz (think of a ring with a rectangular cross section).
     
  8. May 16, 2015 #7
    then its volume should be 2πr.dr.dz.
     
  9. May 16, 2015 #8

    TSny

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    Cylindrical coordinates are fine. But, consider integrating over r first and then z. The upper limit of r will depend on z.
     
  10. May 16, 2015 #9

    TSny

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    I believe you left out something when substituting for ##\cos \phi##. Not that your final expression for dg does not have the correct dimensions for acceleration.
     
  11. May 16, 2015 #10
    Ah yes, my apologies. Think of a chunk of a ring with the ring being cut along the diameter, from two infinitesimally close diameters. Then the volume element is [itex]dV=rdrd\theta dz[/itex]. Basically the r term comes from the Jacobian, which is required in order for the volume element be invariant when performing a coordinate transform [itex](x,y,z)\rightarrow(r,\theta,z)[/itex], so [itex]dV=dxdydz\rightarrow dV'=|J|drd\theta dz=rdrd\theta dz[/itex].
     
  12. May 17, 2015 #11
    sorry. its a typo. i left out the term 'z' in the numerator.
     
    Last edited: May 17, 2015
  13. May 17, 2015 #12
    Do you have picture of this? its difficult to visualize.
     
  14. May 17, 2015 #13
    is this the shape?
    sector.gif
     
  15. May 17, 2015 #14
    @TSny and @kontejnjer so the limits for θ are 0 to 2π
    for r its 0 to ##\sqrt{R^2-z^2}##
    for z its 0 to R.
    So first by integrating with respect to θ, the cunk becomes a full ring. Then integrating with respect to r makes the ring a disc which is situated at z and then integrating with respect to z will make it a hemisphere since it keeps adding one disc over another with changing radius.
     
  16. May 17, 2015 #15
    I also got the correct answer.
    ##g=\pi G\rho R##
     
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