Finding the centre of mass of a hemispherical shell

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Homework Statement


Hey everyone,
I'm studying for my physics and came across a question for the COM of a hemisphere. However, I was wondering what methods there are to find the COM of a hemispherical shell instead. Any insight would be very much appreciated! =)
After much of tearing Google apart in order to find a solution, I found 2, but was wondering if there were any more methods (I always like to find as many solutions as possible because it really helps me understand).

Homework Equations


I just learnt integration in cylindrical and spherical coordinates, so I kinda know that:
[tex]x_\textrm {COM}=\frac{\int x dm}{\int dm} [/tex]
[tex]dA = r^2 \sin\theta\ d\theta\ d\phi [/tex]
[tex]dV = r^2 \sin\theta\ dr\ d\theta\ d\phi [/tex]
where [tex]\theta[/tex] is the polar angle, and [tex]\phi[/tex] is the angle about the equator (I learnt it this way, but I think most people use [tex]\phi[/tex] as the polar angle instead)

The Attempt at a Solution


The 2 methods I found were:
1. Expressing dA as a ratio of the total SA, and this would be equal to the ratio of dm/M; i.e. that [tex]\frac{dA}{2\pi r^2} = \frac{dm}{M}[/tex] (OMG this TeXnology is so cool!!)
ANYWAY so with this, you would be able to
  • find dm
  • find that [itex]z=r \cos \theta[/itex], where z is the axis passing through the apex and centre of the hemisphere.
and plug it all into the equation, giving R/2. YAY!

2. A bit more complicated but really nifty way was to call big radius [itex]R[/itex], and the one on the closer side of the shell [itex]\kappa R[/itex], where [itex]0\leq k\leq 1[/itex].
Then
[tex]\textrm {Total Volume} = \frac{2}{3}\pi R^3 (1-{\kappa}^3)[/tex]
Once again, [itex]z=r \cos \theta[/itex] and [itex]dV = r^2 \sin\theta\ dr\ d\theta\ d\phi \Rightarrow dm = \rho r^2 \sin\theta\ dr\ d\theta\ d\phi[/itex] where [itex]\rho[/itex] is density. From here, you can place into the formula, and end up with
[tex]\frac{3R(1-{\kappa}^4)}{8(1-{\kappa}^3)}[/tex].
Now, as [itex]1-{\kappa}^4 = (1-\kappa)(1+\kappa+{\kappa}^2+{\kappa}^3)[/itex] and [itex]1-{\kappa}^3 = (1-\kappa)(1+\kappa+{\kappa}^2)[/itex], you can simplify this.
For a spherical shell, you just take [tex]\lim_{\substack{x \rightarrow 1}}[/tex], and that gives you R/2 again!!!
 
Last edited:

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