4 
THE KEY. 1ST. M. EEEEEES ON PEOEESSOE SYLYESTEE’S 
Now, by equations (1), we see that 
dm 
dt 
2 /l 1\ 2 , 2 /l 1\ 2 
“ dt~~ C \« 2 b 2 ) ~ ” \c 2 ~ d 2 ) 
=d 2 c\(^-~^p by (3) and (4). 
The other component angular momenta being similarly transformed, we obtain, for 
the component angular momentum about the assigned axis, 
or 
/oij mm 
jus, 
_j/I 1\ 
b 2 y& 2 p^J ‘ c 2 
1 /lao l 
c 2 p 2 
GaW^+V+-5^-pU+V+ c- 
mur 
(<) 
Comparing this with the expression already obtained for the angular momentum of the 
body, we see that the two expressions are equal if 
l“> 1 , . nW 3_(\ 
pi + c 2 — 
?. e. if the axis be parallel to the rough plane ; and generally that the angular momentum 
of the ellipsoid, that of the particle, 
=(G— H )^(^+’ 
! + : 
which, if the axis be perpendicular to the rough plane, becomes (G — H) ~~r~ a con- 
stant. — February 1870]. 
Now, let Ji t denote the angular momentum of the ellipsoid about an axis through its 
centre, parallel to the rough plane, and at right angles to the instantaneous axis, and h 
about an axis through its centre parallel to the rough plane and at right angles to this 
last. The radius vector of the point of contact measured from the foot of the perpen- 
dicular on the rough plane is - (m ; and hence if n denote the angular velocity of this 
radius vector in space, the radial and transversal component velocities of the point of 
p dfj* p 
contact will be y - nuj respectively. To obtain the component angular momenta 
of the particle we have only to multiply the expressions by G. Hence, by the theorem 
just proved, 
,;F dp 
X dt 
( 8 ) 
We can now calculate the values of P and F, the pressure and friction of the ellipsoid 
against the plane. Taking moments about axes through the centre of the ellipsoid, 
lying in the rough plane and perpendicular to the direction of P and F respectively, 
