A spherical shell of mass M and radius R is completely filled with a frictionless fluid, also of mass M. It is released
from rest, and then it rolls without slipping down an incline that makes an angle θ with the horizontal. What will
be the acceleration of the shell down the incline just after it is released? Assume the acceleration of free fall is g.
The moment of inertia of a thin shell of radius r and mass m about the center of mass is I =2/ 3mr2; the moment
of inertia of a solid sphere of radius r and mass m about the center of mass is I =2/5mr2.
(A) a = g sin θ
(B) a =3/4 g sin θ
(C) a =1/2 g sin θ
(D) a =3/8 g sin θ
(E) a =3/5 g sin θ
The Attempt at a Solution
I considered the fluid inside to be a "sphere in sphere" and used conservation of energy to get 2M*g*h=M*v^2 + 2/5*Mv^2+ 2/3 * Mv^2. (I used the fact that 2*1/2 =1 and that v=omega*r to simply). Adding the right hand side and diving by M give me 2*g*h= 29/15 v^2. I then divide by 2*h to get g=(29/15*v^2)/2h. How do I go from here?