We have
#l=sqrt(24)a#
#l_g=l+1/5 l = 6/5l#
#vec r_g = (0,l_g/cosalpha,0)#
where #(u,v,w)# must be understood as
#u hat i + v hat j + w hat k#
Here
#alpha = arctan(1/sqrt(24))#
#vec omega = (0,cosalpha,sinalpha)omega#
#vec v_g = vec r_g xx vec omega = 17/2a^2m omega(0,cosalpha,sinalpha)#
then
#abs(vec v_g)=17/2a^2m omega#
and
(D) #Omega = abs(vec v_g)/abs(vec r_g) = 1/5 a omega#
or
#vec Omega = (0,0,1)Omega#
Now
#vec L_O = J_(omega) vec omega+J_(Omega) vec Omega#
with
#J_(omega)=(ma^2)/2+(4m(2a)^2)/2#
We have also
#vec L_g = J_(omega) vec omega# and
(B) #norm(vec L_g)=17/2a^2m omega#
and
#J_(Omega)=(ma^2)/4+ml_0^2+((4m)(2a)^2)/4+4m(2l_0)^2#
with #l_0 = l cos alpha#
so
(A) #<< vec L_O, hat k >> = J_(Omega)Omega+J_(omega)<< vec omega, hat k >> = J_(Omega)Omega+J_(omega)omega sin alpha#
and
#norm(vec L_O) = sqrt(<< vec L_O, hat k >>^2+(J_(omega)omega cosalpha)^2 )#
or
(C) #norm(vec L_O) = sqrt((J_(Omega)Omega)^2+2J_(Omega)J_(omega) Omega omega sinalpha+(J_(omega)omega)^2)#
The final numeric results are left to the reader as an exercise.
