Centre of mass of a solid hemisphere.

[grahammtb]

Hi there, I can't get my head round how to do the math for this problem. I'm sure it's not as hard as I think...

Homework Statement

Show that the CoM of a uniform solid hemisphere of radius r lies at a distance (3/8)r from the centre of the flat face.
You may find it convenient to regard the hemisphere as consisting of a very large number of extremely thin discs of varying radii stacked on top of each other.

Homework Equations

I'm thinking I'll need the equation for CoM, involving an integral: R=(1/M)\intrdm. This I think can be modified to: R=(1/M)\intr\rhodV, where rho is the density at radius r.

The Attempt at a Solution

N/A

Thanks very much for any help!
~Graham

 

[Doc Al]

I'm thinking I'll need the equation for CoM, involving an integral: R=(1/M)\intrdm. This I think can be modified to: R=(1/M)\intr\rhodV, where rho is the density at radius r.

So far, so good. To avoid confusion, I recommend you use a different variable of integration than r--let's say y. (Imagine the hemisphere axis to be along the y-axis from y = 0 to y = r.) What's the volume of a thin disk located at position y?

 

[grahammtb]

So far, so good. To avoid confusion, I recommend you use a different variable of integration than r--let's say y. (Imagine the hemisphere axis to be along the y-axis from y = 0 to y = r.) What's the volume of a thin disk located at position y?

Ok, since the discs are extremely thin, I'd approximate the volume of the disc to be its area. So: A(y) = Pi.(r(y))².

 

[tiny-tim]

Hi Graham!

… You may find it convenient to regard the hemisphere as consisting of a very large number of extremely thin discs of varying radii stacked on top of each other.
…
… dV …

You're ignoring the hint …

if you use discs stacked on top of each other (with height z, say), you can go straight to integrating over dz, instead of dV.

 

[Doc Al]

Ok, since the discs are extremely thin, I'd approximate the volume of the disc to be its area. So: A(y) = Pi.(r(y))².

The disks are thin, but not zero thickness! Hint: dV = Area dy.

 

[grahammtb]

The disks are thin, but not zero thickness! Hint: dV = Area dy.

Ah I think I see...so if dV = Ady, then I can put that into my original eqn. for CoM? R = (1/M)\int\rhoydV. Then R = (Pi.\rho/M)\inty³dy...I think

 

[Doc Al]

Ah I think I see...so if dV = Ady, then I can put that into my original eqn. for CoM? R = (1/M)\int\rhoydV. Then R = (Pi.\rho/M)\inty³dy...I think

You're on the right track, but not quite there yet. What's the radius of a disk at position y? (Draw yourself a diagram.)

 

[grahammtb]

yep! It worked! Thanks a lot guys, I'd have spent all afternoon trying to pick my way through :)