Masoud Masoud

Posted on Sunday, 09 March, 2003 - 11:19 am:

A uniform solid consists of a hemisphere of radius r and a right circular cone of base radius r and height h, fixed together so that their plane faces coincide. The solid can rest in equilibrium with any point ot the curved surface area in contact with a horizontal plane. Find h in terms of r.

I know that in this question the centre of mass must be assumed to be on the part of the solid where both circular faces meet however I don not know how to get h in terms of r, what's the strategy?

Thank you

Mark Durkee

Posted on Sunday, 09 March, 2003 - 11:34 am:

Say that the sum of the moments of the two components of the solid about the line of intersection is zero.

Therefore:

$π r^{2} h / 3 \times h / 4 = 2 π r^{3} / 3 \times 3 r / 8$

Mark Durkee

Posted on Sunday, 09 March, 2003 - 11:58 am:

I've just realised that i didn't explain where the parts of the equation come from.

$π r^{2} h / 3$ is the volume of the cone (proportional to its mass and therefore its weight)

$h / 4$ is the distance of the COM of a solid cone from its base

$2 π r^{3} / 3$ is the volume of the hemisphere

$3 r / 8$ is the distance of the COM of the hemisphere from the flat surface
So for the sum of the moments to be zero about the line of intersection:
Volume of cone x Distance of COM of cone from the line of intersection = Volume of hemishpere x Distance of COM of hemisphere from line of intersection

Masoud Masoud

Posted on Sunday, 09 March, 2003 - 02:25 pm:

Thank you for taking the time to answer my question.