# Hemisphere angle of tilt where centre of mass near the edge mechanics help required

• May 23rd 2009, 07:13 PM
Hemisphere angle of tilt where centre of mass near the edge mechanics help required
The hemisphere of radius a has centre of mass at 3/2a, 1/6 a as shown in the diagram. How do I find the angle between PC and horizontal when the hemisphere is in equilibrium with its curved surface on ground?
http://www.mathhelpforum.com/math-he...1&d=1243134758
• May 23rd 2009, 09:13 PM
chisigma
The equilibrium condition is equivalent to minimum energy condition and that means that the ordinate of center of mass must be the lowest possibile, so that the angle between PC and horizontal [in radians] must be...

$\displaystyle \theta = tan^{-1} \frac {\frac{a}{2}}{\frac{a}{6}}= tan^{-1} 3 = 1,24904577...$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 23rd 2009, 10:33 PM
Quote:

Originally Posted by chisigma
The equilibrium condition is equivalent to minimum energy condition and that means that the ordinate of center of mass must be the lowest possibile, so that the angle between PC and horizontal [in radians] must be...

$\displaystyle \theta = tan^{-1} \frac {\frac{a}{2}}{\frac{a}{6}}= tan^{-1} 3 = 1,24904577...$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

can you please draw a picture of this? I am having some trouble imagining it.(Wondering)
• May 24th 2009, 03:49 AM
chisigma
The equilibrium condition is also the 'minimum energy' condition and it is realized when the center of mass is at the minimum possible quote. The value of $\displaystyle \theta$ I have indicated minimizes the quote of the center of mass, that means that for any other value of $\displaystyle \theta$ the 'potential energy' is greater...

http://digilander.libero.it/luposabatini/MHF17.bmp

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$