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Center of mass of a hemisphere

  1. Sep 6, 2011 #1
    As part of a problem I need to find the center of mass of a hollow hemisphere.

    I did it in the following way.

    I considered the lower hemisphere ( around negative z axis )

    First by definition, the position vector of the center of mass is obtained as

    [itex]\vec{R}[/itex]= [itex]\frac{\int\vec{r}dm}{\int dm}[/itex]


    Now I set up the problem in spherical coordinates

    http://img855.imageshack.us/img855/4741/spherev.png [Broken]

    Uploaded with ImageShack.us


    Now, [itex]\vec{r}[/itex] = R[itex]\hat{r}[/itex]

    where [itex]\hat{r}[/itex] = sin[itex]\theta[/itex]cos[itex]\phi[/itex][itex]\hat{i}[/itex]+sin[itex]\theta[/itex]sin[itex]\phi[/itex][itex]\hat{j}[/itex]+cos[itex]\theta[/itex][itex]\hat{k}[/itex]

    And dm = [itex]\sigma[/itex]ds, where ds is the area element, and [itex]\sigma[/itex] is the mass per unit area.

    ds = R2sin[itex]\theta[/itex]d[itex]\theta[/itex]d[itex]\phi[/itex]

    So then [itex]\vec{R}[/itex]= [itex]\frac{\int\vec{r}ds}{\int ds}[/itex]

    i.e (2[itex]\pi[/itex]R2) [itex]\vec{R}[/itex]=[itex]\int[/itex][itex]\int[/itex]R3 (sin2[itex]\theta[/itex]cos[itex]\phi[/itex][itex]\hat{i}[/itex]+sin2[itex]\theta[/itex]sin[itex]\phi[/itex][itex]\hat{j}[/itex]+[itex]\frac{sin2\theta}{2}[/itex][itex]\hat{k}[/itex])d[itex]\theta[/itex]d[itex]\phi[/itex]


    integrating [itex]\theta[/itex] from [itex]\pi[/itex]/2 to [itex]\pi[/itex] and [itex]\phi[/itex] from 0 to 2[itex]\pi[/itex]

    I get [itex]\vec{R}[/itex] = -[itex]\frac{\hat{k}}{2}[/itex]

    now I know this is not the right answer, the correct answer is -[itex]\frac{2\pi}{R}[/itex][itex]\hat{k}[/itex]

    I know that the problem can be simplified with symmetry and done with much more ease than what I have attempted, but I'd like to know what is wrong with the way I formulated the problem.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 6, 2011 #2

    vela

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    This is the correct answer.
    This can't possible be the correct answer. The units aren't correct.
     
  4. Sep 6, 2011 #3

    sorry I made a mistake, the correct answer is- [itex]\frac{2R}{\pi}[/itex][itex]\hat{K}[/itex]


    You can arrive at that answer with this argument, the sphere can be thought of to be made of a number of semicircular line segments of radius R ( or taking one and rotating it about the negative Z axis )

    So the C.M of the sphere will be the same as one such segment, since the rotation occurs about an axis through the center of mass.

    http://img97.imageshack.us/img97/3853/semit.png [Broken]

    Uploaded with ImageShack.us


    now, if I consider a horizontal strip, the CM of the strip is on the Z axis at a distance Rcos[itex]\theta[/itex] and the mass dm = 2[itex]\lambda[/itex]dl , where [itex]\lambda[/itex] is the linear mass density, [itex]\lambda[/itex] gets canceled out and the cm is at

    [itex]\vec{R}[/itex] =[itex]\frac{1}{\pi R}[/itex][itex]\int[/itex]2R2cos[itex]\theta[/itex]d[itex]\theta[/itex]


    dl = Rd[itex]\theta[/itex]

    [itex]\theta[/itex] varies from 0 to [itex]\frac{\pi}{2}[/itex]


    then [itex]\vec{R}[/itex] = [itex]\frac{2R}{\pi}[/itex]
     
    Last edited by a moderator: May 5, 2017
  5. Sep 6, 2011 #4

    vela

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    The radius of the circular strip isn't R. It's only equal to R when theta is pi/2.
     
  6. Sep 6, 2011 #5
    you misunderstand what i mean, I am taking a semicircle of radius R and rotating it , think of it like an electric drill with a leaf as a semicircle,

    when it spins it will form a hemisphere right?

    P.S

    This is not what is happening in the integral, i just used the notion of rotation to infer that the CM of both the semicircle and hemisphere have to be the same
     
  7. Sep 6, 2011 #6

    vela

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    Your initial assumption that you can break the sphere up into a bunch of line segments of constant density is flawed. Think about breaking the sphere up into a finite number of wedges. The wedge is thicker at the top than at the bottom, so there's more mass near the top than at the bottom. This doesn't change as the number of wedges goes to infinity. So the line density should have a factor of sin θ.
     
  8. Sep 6, 2011 #7
    I don't follow, since this is a hollow sphere, what do you mean by wedges?
     
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